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Valentin BojuLouis Funar
The Math Problems Notebook
BirkhauserBoston • Basel • Berlin
Valentin BojuMontrealTechInstitut de Technologie de MontrealP.O. Box 78575, Station WildertonMontreal, Quebec, H3S 2W9 [email protected]
Louis FunarInstitut Fourier BP 74URF MathematiquesUniversite de Grenoble I38402 Saint Martin d’Heres [email protected]
Cover design by Alex Gerasev.
Mathematics Subject Classification (2000): 00A07 (Primary); 05-01, 11-01, 26-01, 51-01, 52-01
Library of Congress Control Number: 2007929628
ISBN-13: 978-0-8176-4546-5 e-ISBN-13: 978-0-8176-4547-2
Printed on acid-free paper.
c©2007 Birkhauser BostonAll rights reserved. This work may not be translated or copied in whole or in part without the writtenpermission of the publisher (Birkhauser Boston, c/o Springer Science+Business Media LLC, 233 SpringStreet, New York, NY 10013, USA), except for brief excerpts in connection with reviews or scholarlyanalysis. Use in connection with any form of information storage and retrieval, electronic adaptation,computer software, or by similar or dissimilar methodology now known or hereafter developed is forbid-den.The use in this publication of trade names, trademarks, service marks and similar terms, even if they arenot identified as such, is not to be taken as an expression of opinion as to whether or not they are subjectto proprietary rights.
9 8 7 6 5 4 3 2 1
www.birkhauser.com (TXQ/SB)
The Authors
Valentin Boju was professor of mathematics at the University of Craiova, Romania,until his retirement in 2000. His research work was primarily in the field of geome-try. He further promoted biostatistics and biomathematics as a discipline within theDepartment of Medicine of the University of Craiova. He left Romania for Canada,where he has taught at MontrealTech since 2001. He was actively involved in coach-ing college students in problem-solving and intuitive mathematics. In 2004, he washonored with the title of Officer of the Order “Cultural Merit,” Category “ScientificResearch.”
MontrealTechInstitut de Technologie de MontréalP.O. Box 78575 Station WildertonMontréal, Quebec, H3S 2W9Canadae-mail: [email protected]
Louis Funar has been a researcher at CNRS, at the Fourier Institute, University ofGrenoble, France, since 1994. During his college years, he participated three timesin the International Mathematics Olympiads with the Romanian team in the years1983–1985, winning bronze, gold, and silver medals. His main research interests arein geometric topology and mathematical physics.
Institut Fourier, BP 74Mathématiques UMR 5582Université de Grenoble I38402 Saint-Martin-d’Hères cedexFrancee-mail: [email protected]
To all schoolteachers, particularly to my wonderful parents,Evdochia and Nicolae Boju, who genuinely believed in theeducation of the younger generation.
To my parents, Maria and Ioan Funar, who laid the foundationfor our collaboration, as a late reward for their hard work andhelp, for their enthusiasm, determinedness, and strengthduring all these years.
Preface
The authors met on a Sunday morning about 25 years ago in Room 113. One of uswas a college student and the other was leading the Sunday Math Circle. This circle,within the math department of the University of Craiova, gathered college studentswho possessed a common passion for mathematics. Most of them were participatingin the mathematical competitions in vogue at that time, namely the Olympiads. Othersjust wanted to have good time.
There were similar math circles everywhere in the country, in which high-schoolstudents were committed to active training for math competitions. The highest stan-dard was achieved by the selective training camps organized at the national level,which were led by professional coaches and mathematicians who were able to stimu-late and elicit high performance. To mention a few among the leaders, we recall DorinAndrica, TomaAlbu, TituAndreescu, Mircea Becheanu, Ioan Cuculescu, Dorel Mihet,and Ioan Tomescu.
The Sunday Math Circle shifted partially from its purpose of Olympiad trainingby following freely the leader’s ideas and thus becoming a place primarily intendedfor disseminating beautiful mathematics at an elementary level. The fundamentaltexts were the celebrated book of Richard Courant and Herbert Robbins with themysterious title What is mathematics?, and the book of David Hilbert and StefanCohn–Vossen entitled Geometry and the Imagination. The participants soon realizedthe differences in both scale and novelty between the competition-type math problemsthat they encountered every day and the mathematics built up over hundreds of yearsand engaged in by professional mathematicians.
Competition problems have to be solved in a short amount of time, and peoplecompete against each other. These might be highly nontrivial, unconventional prob-lems requiring deep insights and a lot of imagination, but still they have the advantagethat one knows in advance that they have a solution. In the real mathematical world,problems often had to wait not merely years but sometimes centuries to be solved. Aworking mathematician could go for months or years trying to solve a particular prob-lem that had not been solved before and take the risk of bitterly failing. Moreover, thesustained effort needed for accomplishing such a task would have to take into accounta delicate balance between the accumulation of knowledge, methods, and tools and
viii Preface
the creation from scratch of a path leading to a solution. This intellectual adventureis filled with suspense and frustration, since one does not know for sure what one istrying to prove or whether it is indeed true.
Real mathematics seemed to many of us to be too far away from competitionproblems. The philosophy of the Sunday Math Circle was that, in contrast to what wemight think, the border between the two kinds of mathematics could be so vague andflexible that at times one could cross it at an early age. The history of mathematicsabounds in examples in which a fresh mind was able to find an unexpected solutionthat specialists had been unable to find. One needs, however, the right training and theunconventional sense of finding the inspiring problem. Eventually, once a solutionhas been found, mathematicians are then willing to try to understand it even better,and other solutions follow in time, each one simpler and clearer than the previousone. In some sense, once solved, even the hardest problems start losing slowly theiraura of difficulty and eventually become just problems. Problems that are today partof the curriculum of the average high-school student were difficult research problemsthree hundred years ago, solved only by brilliant mathematicians. This phenomenondemonstrates the evolution that language and science have undergone since then.
The authors conceived the present book with the nostalgia of the “good old times"of the Sunday Math Circles. We wanted something that carries the mark of that philos-ophy, namely, a number of challenging math problems for Olympiads with a glimpseof related problems of interest for the mathematician. The present text is a collectionof problems that we think will be useful in training students for mathematical com-petitions. On the other hand, we hope that it might fulfill our second goal, namely,that of awakening interest in advanced mathematics. Thus its audience might rangefrom college students and teachers to advanced math students and mathematicians.
The problems in each section are in increasing order of difficulty, so that the readergive some of the problems a try. We wanted to have 25% easy problems concerningbasic tools and methods and consisting mainly of instructional exercises. The beginnermight jump directly to the solutions, where a concentrate of the general theory andbasic tricks can be found. The largest chunk contains about 50% problems of mediumdifficulty, which could be useful in training for mathematical competitions from localto international levels. The remaining 25% might be considered difficult problemseven for the experienced problem-solver. These problems are often accompanied bycomments that put the results in a broader perspective and might incite the reader topursue the research further.
The problems serve as an excuse for introducing all sorts of generalizations andclosely related open problems, which are spread among the solutions. Some of theseare truly outstanding problems that resisted the efforts of mathematicians over thecenturies, such as the congruent numbers conjecture and the Riemann hypothesis,which are among seven Millennium Prize Problems that the Clay Mathematics Insti-tute recorded as some of the most difficult issues with which mathematicians werestruggling at the turn of the second millennium and offered a reward of one milliondollars for a solution to each one. In mathematics the frontier of our knowledge isstill open, and it abounds in important unsolved problems, many of which can be
Preface ix
understood at the undergraduate level. The reader might be soon driven to the edgeof that part of mathematics where he or she could undertake original research.
We drew inspiration from the spirit of the famous books by R. K. Guy and hiscollaborators. Nevertheless, our aim was not to build up a collection of open questionsin elementary mathematics but rather to offer a journey among the basic methods inproblem-solving. Developing intuition and strengthening the most popular techniquesin mathematical competitions are equally part of the goal. In the meantime, we offer theenthusiastic reader a brief glimpse of mathematical research, a place where problemsyet unsolved long for deliverance.
The present collection of problems evolved from a notebook in which the second-named author collected the most interesting and unconventional problems that heencountered during his training for mathematical competitions in the 1980s. In thetradition of the Romanian school of mathematical training, he encountered problemsinspired by both Russian Olympiads and American Competitions. Some (if not most)of the problems have already appeared elsewhere, especially in Kvant, Matematikav Shkole, American Mathematical Monthly, Elemente der Mathematik, MatematikaiLapok, Gazeta Matematica, and so, in a sense, the collection gained a certain cos-mopolitan flavor. We have given the source in the problem section, when known, andmore detailed information in the comments within the solutions part.
The original set of problems is complemented by more basic exercises, whichaim at introducing many of the tricks and methods of which the competitor should beaware. Years later, we followed the destiny of some of the most intriguing questionsfrom the notebook. Some of them led to developments that are well beyond the scopeof this book, and we decided to outline a few of them.
We have supplied a short glossary containing some less usual definitions andidentities in the geometry of triangles and the solution of Pell’s equation. In order tohelp the reader find his or her way through the book, we have provided both an indexconcerning all mathematical results needed in the proofs, which are usually stated atthe place where they are used first, and an index of mathematical terms at the end ofthe book.
The authors have benefited from discussions, corrections, help, and feedback fromseveral people, whom we want to thank warmly: Dorin Andrica, Barbu Berceanu,Roland Bacher, Maxime Wolff, Mugurel Barcau, Ioan Filip, and Simon GyörgySzatmari. We also thank Ann Kostant, Editorial Director, Springer, and AvantiParanjpye, Associate Editor, Birkhäuser Boston, for their productive suggestions.One of them led to the “Index of Topics and Methods,” which might be useful fora better reception of the book by readers. We are very grateful to Elizabeth Loew,our Production Editor, for her patience and dedication to accuracy and excellence.Finally, we are thankful to David Kramer for his thorough copyediting corrections.
Valentin Boju and Louis FunarMontreal, Canada and Saint-Martin-d’Hères, FranceJuly 2006
Contents
Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vii
Part I Problems
1 Number Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
2 Algebra and Combinatorics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112.1 Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112.2 Algebraic Combinatorics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.3 Geometric Combinatorics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
3 Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213.1 Synthetic Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213.2 Combinatorial Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233.3 Geometric Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
4 Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
Part II Solutions and Comments to the Problems
5 Number Theory Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
6 Algebra and Combinatorics Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . 856.1 Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 856.2 Algebraic Combinatorics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 956.3 Geometric Combinatorics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120
7 Geometry Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1337.1 Synthetic Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1337.2 Combinatorial Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142
xii Contents
7.3 Geometric Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175
8 Analysis Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185
9 Glossary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2179.1 Compendium of Triangle Basic Terminology and Formulas . . . . . . . 217
9.1.1 Lengths in a Triangle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2179.1.2 Important Points and Lines in a Triangle . . . . . . . . . . . . . . . . . 2189.1.3 Coordinates in a Triangle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219
9.2 Appendix: Pell’s Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 220
Index of Mathematical Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225
Index of Mathematical Terms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 227
Index of Topics and Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231
Part I
Problems
1
Number Theory
Problem 1.1. Show that we have Ckn ≡ 0 (mod 2) for all k satisfying 1 ≤ k ≤ n− 1if and only if n = 2β , where β ∈ Z
∗+. Here Ckn denotes the number of combinations,i.e., the number of ways of picking up a subset of k elements from a set of n elements.Known also as the binomial coefficient or choice number and sometimes denoted as(nk
)it is given by the formula
Ckn =(n
k
)= n!
k!(n− k)!
where the factorial n! represents
n! = 1 × 2 × 3 × · · · × n.
Problem 1.2. Let P = anxn + · · · + a0 be a polynomial with integer coefficients.
Suppose that there exists a number p such that:
1. p does not divide an;2. p divides ai , for all i ≤ n− 1;3. p2 does not divide a0.
Then P is an irreducible polynomial in Z[x].
Problem 1.3. Given mi, bi ∈ Z+, i ∈ {1, 2, . . . , n}, such that gcd(mi,mj ) = 1 forall i �= j , there exist integers x satisfying x ≡ bi (mod mi) for all i. This result isusually known as the Chinese remainder theorem.
Problem 1.4. If gcd(a,m) = 1, then it is a classical result of Euler that we have thefollowing congruence:
aϕ(m)+1 ≡ 0 (mod m),
where ϕ(m) is the Euler totient function, which counts how many positive integerssmaller than m are relatively prime to m. Prove that this equality holds precisely forthose numbers a,m such that for any prime number p that divides a, if pk divides mthen pk also divides a.
4 1 Number Theory
Problem 1.5. (Kvant) Let p > 2 be a prime number and ak ∈ {0, 1, . . . , p2 − 1}denote the value of kp modulo p2. Prove that
p−1∑
k=1
ak = p3 − p2
2.
Problem 1.6. If k ∈ Z+, then show that[√k2 + 1 + · · · +
√k2 + 2k
]= 2k2 + 2k,
where [x] denotes the integer part, i.e., the largest integer smaller than x.
Problem 1.7. (Amer. Math. Monthly) If n is not a multiple of 5, then P = x4n +x3n + x2n + xn + 1 is divisible by Q = x4 + x3 + x2 + x + 1.
Problem 1.8. (Amer. Math. Monthly) Let p be an odd prime and k ∈ Z+. Show thatthere exists a perfect square the last k digits of whose expansion in base p are 1.
Problem 1.9. Any natural number greater than 6 can be written as a sum of twonumbers that are relatively prime.
Problem 1.10. (International Math. Olympiad) Prove that it is impossible to extractan infinite arithmetic progression from the sequence S = {1, 2k, 3k, . . . , nk, . . . },where k ≥ 2.
Problem 1.11. (Amer. Math. Monthly) Prove that ba−j+1 dividesCjba if a, b ≥ 2, j ≤a + 1.
Problem 1.12. Solve in integers the following equations: (1) x2 = y2 + y3; (2)x2 + y2 = z2.
Problem 1.13. Let a, b, c, d ∈ Z+ be such that at least one of a and c is not a perfectsquare and gcd(a, c) = 1. Show that there exist infinitely many natural numbers nsuch that an + b, cn + d are simultaneously perfect squares if one of the followingconditions is satisfied:
1. b and d are perfect squares;2. a + b, c + d are perfect squares;3. a(d − 1) = c(b − 1).
Moreover, there do not exist such numbers if a = 1, b = 0, c = 4k2 − 1, d = 1.
Problem 1.14. If N = 2 + 2√
28n2 + 1 ∈ Z for a natural number n, then N is aperfect square.
Problem 1.15. Let n ≥ 5, 2 ≤ b ≤ n. Prove that[(n− 1)!
b
]≡ 0 (mod b − 1).
1 Number Theory 5
Problem 1.16. (Amer. Math. Monthly) Prove that for every natural number n, thereexists a natural number k such that k appears in exactly n nontrivial Pythagoreantriples.
Problem 1.17. (Amer. Math. Monthly) Let n, q ∈ Z+ be such that all prime divisorsof q are greater than n. Show that
(q − 1)(q2 − 1) · · · (qn−1 − 1) ≡ 0 (mod n!).
Problem 1.18. Every natural number n ≥ 6 can be written as a sum of distinct primes.
Problem 1.19. Every number n ≥ 6 can be written as a sum of three numbers thatare pairwise relatively prime.
Problem 1.20. (Romanian training camp, Sinaia 1984) Find all pairs of integers(m, n) such that
Cnm = 1984,
where Cnm = m!n!(m−n)! denotes the usual binomial coefficient.
Problem 1.21. (Romanian training camp, Sinaia 1984) Find the set A consisting ofnatural numbers n that are divisible by all odd natural numbers a with a2 < n.
Problem 1.22. Prove that 21092 − 1 is divisible by 10932.
Problem 1.23. (Amer. Math. Monthly) Let n ≥ 0, r > 1, and 0 < a ≤ r be threeintegers. Prove that the number n, when written in base r , has precisely
∞∑
k=1
[nr−k + ar−1
]−
[nr−k
]
digits that are greater than or equal to r − a.
Problem 1.24. Find a pair (a, b) of natural numbers satisfying the following proper-ties:
1. ab(a + b) is not divisible by 7.2. (a + b)7 − a7 − b7 is divisible by 77.
Problem 1.25. (International Math. Olympiad 1984) Let 0 < a < b < c < d be oddintegers such that
1. ad = bc
2. a + d = 2k, b + c = 2m, for some integers k and m.
Prove that a = 1.
Problem 1.26. (Amer. Math. Monthly) Find those subsets S ⊂ Z+ such that all butfinitely many sums of elements from S (possibly with repetitions) are compositenumbers.
6 1 Number Theory
Problem 1.27. (Amer. Math. Monthly) Prove that for any natural n > 1, the number2n − 1 does not divide 3n − 1.
Problem 1.28. (Putnam Competition 1964) Let un be the least common multiple ofthe first n terms of a strictly increasing sequence of positive integers. Prove that
∞∑
n=1
1
un≤ 2.
Find a sequence for which equality holds above.
Problem 1.29. (Amer. Math. Monthly) Let ϕn(m) = ϕ(ϕn−1(m)), where ϕ1(m) =ϕ(m) is the Euler totient function, and set ω(m) the smallest number n such thatϕn(m) = 1. If m < 2α , then prove that ω(m) ≤ α.
Problem 1.30. Let f be a polynomial with integer coefficients andN(f ) = card{k ∈Z; f (k) = ±1}. Prove thatN(f ) ≤ 2 + deg f , where deg f denotes the degree of f .
Problem 1.31. Prove that every integer can be written as a sum of 5 perfect cubes.
Problem 1.32. If n ∈ Z, then the binomial coefficient C2n = (2n)!(n!)2
has an evennumber of divisors.
Problem 1.33. (Amer. Math. Monthly) Prove that every n ∈ Z+ can be written inprecisely k(n) different ways as a sum of consecutive integers, where k(n) is thenumber of odd divisors of n greater than 1.
Problem 1.34. (Amer. Math. Monthly) Let π2(x) denote the number of twin primesp with p ≤ x. Recall that p is a twin prime if both p and p+ 2 are prime. Show that
π2(x) = 2 +∑
7≤n≤xsin
(π
2(n+ 2)
[n!
n+ 2
])sin
(π
2n
[(n− 2)!
n
])
for x > 7.
Problem 1.35. (Kvant)
1. Find all solutions of the equation 3x+1 + 100 = 7x−1.2. Find two solutions of the equation 3x + 3x
2 = 2x + 4x2, and prove that there are
no others.
Problem 1.36. (Kvant) Let σ(n) denote the sum of the divisors of n. Prove that thereexist infinitely many integers n such that σ(n) > 2n, or even stronger, such thatσ(n) > 3n. Prove also that σ(n) < n(1 + log n).
Problem 1.37. (American Competition) Let ai be natural numbers such thatgcd(ai, aj ) = 1 and the ai are not prime numbers. Show that
1
a1+ · · · + 1
an< 2.
1 Number Theory 7
Problem 1.38. Let σ(n) denote the sum of divisors of n. Show that σ(n) = 2k if andonly if n is a product of Mersenne primes, i.e., primes of the form 2k − 1, for k ∈ Z.
Problem 1.39. (Putnam Competition) Find all integer solutions of the equation |pr −qs | = 1, where p, q are primes and r, s ∈ Z \ {0, 1}.Problem 1.40. Consider an arithmetic progression with ratio between 1 and 2000.Show that the progression does not contain more than 10 consecutive primes.
Problem 1.41. (Amer. Math. Monthly) Let a1 = 1, an+1 = an + [√an
]. Show that
an is a perfect square iff n is of the form 2k + k − 2.
Problem 1.42. Recall that ϕ(n) denotes the Euler totient function (i.e., the numberof natural numbers less than n and prime to n), and that σ(n) is the sum of divisorsof n. Show that n is prime iff ϕ(n) divides n− 1 and n+ 1 divides σ(n).
Problem 1.43. (Amer. Math. Monthly) A number is called ϕ-subadditive if ϕ(n) ≤ϕ(k) + ϕ(n − k) for all k such that 1 ≤ k ≤ n − 1, and ϕ-superadditive if thereverse inequality holds. Prove that there are infinitely many ϕ-subadditive numbersand infinitely many ϕ-superadditive numbers.
Problem 1.44. (Amer. Math. Monthly) Find the positive integers N such that for alln ≥ N , we have ϕ(n) ≤ ϕ(N).
Problem 1.45. A number n is perfect if σ(n) = 2n, where σ(n) denotes the sum of alldivisors of n. Prove that the even number n is perfect if and only if n = 2p−1(2p−1),where p is a prime number with the property that 2p − 1 is prime.
Problem 1.46. (Elemente der Mathematik) A number n is superperfect if σ(σ(n)) =2n, where σ(k) is the sum of all divisors of k. Prove that the even number n issuperperfect if and only if n = 2r , where r is an integer such that 2r+1 − 1 is prime.
Problem 1.47. Ifa, b are rational numbers satisfying tan aπ = b, thenb ∈ {−1, 0, 1}.Problem 1.48. (Amer. Math. Monthly) LetAn = run+ svn, n ∈ Z+, where r, s, u, vare integers,u �= ±v, and letPn be the set of prime divisors ofAn. ThenP = ⋃∞
n=0 Pnis infinite.
Problem 1.49. Solve in natural numbers the equation
x2 + y2 + z2 = 2xyz.
Problem 1.50. (Amer. Math. Monthly) Find the greatest common divisor of the fol-lowing numbers: C1
2n, C32n, C
52n, . . . , C
2n−12n .
Problem 1.51. (Amer. Math. Monthly) Prove that if Sk = ∑ni=1 k
gcd(i,n), then Sk ≡0 (mod n).
8 1 Number Theory
Problem 1.52. Prove that for any integer n such that 4 ≤ n ≤ 1000, the equation
1
x+ 1
y+ 1
z= 4
n
has solutions in natural numbers.
Problem 1.53. (Nieuw Archief v. Wiskunde) For a natural number n we set S(n) forthe set of integers that can be written in the form 1+g+· · ·+gn−1 for some g ∈ Z+,g ≥ 2.
1. Prove that S(3) ∩ S(4) = ∅.2. Find S(3) ∩ S(5).
Problem 1.54. If k ≥ 202 and n ≥ 2k, then prove that Ckn > nπ(k), where π(k)denotes the number of prime numbers smaller than k.
Problem 1.55. (Amer. Math. Monthly) Let Sm(n) be the sum of the inverses of theintegers smaller thanm and relatively prime to n. Ifm > n ≥ 2, then show that Sm(n)is not an integer.
Problem 1.56. Solve in integers the following equations:
1. x4 + y4 = z2;2. y3 + 4y = z2;3. x4 = y4 + z2.
Problem 1.57. Show that the equation
x3 + y3 = z3 + w3
has infinitely many integer solutions. Prove that 1 can be written in infinitely manyways as a sum of three cubes.
Problem 1.58. Prove that the equation y2 = Dx4 + 1 has no integer solution exceptx = 0 if D �≡ 0,−1, 3, 8 (mod 16) and there is no factorization D = pq, wherep > 1 is odd, gcd(p, q) = 1, and either p ≡ ±1 (mod 16), p ≡ q ± 1 (mod 16), orp ≡ 4q ± 1 (mod 16).
Problem 1.59. (Amer. Math. Monthly) Find all numbers that are simultaneously tri-angular, perfect squares, and pentagonal numbers.
Problem 1.60. (Amer. Math. Monthly) Find all inscribable integer-sided quadrilater-als whose areas equal their perimeters.
Problem 1.61. (Amer. Math. Monthly) Every rational number a can be written as asum of the cubes of three rational numbers. Moreover, if a > 0, then the three cubescan be chosen to be positive.
Problem 1.62. 1. Every prime number of the form 4m + 1 can be written as thesum of two perfect squares.
1 Number Theory 9
2. Every natural number is the sum of four perfect squares.
Problem 1.63. Every natural number can be written as a sum of at most 53 integersto the fourth power.
Problem 1.64. Let G(k) denote the minimal integer n such that any positive integercan be written as the sum of n positive perfect kth powers. Prove that G(k) ≥ 2k +[
3k
2k
]− 2.
Problem 1.65. (Amer. Math. Monthly) LetR(n, k) be the remainder whenn is dividedby k and
S(n, k) =k∑
i=1
R(n, i).
1. Prove that limn→∞ S(n, n)
n2 = 1 − π2
12 .2. Consider a sequence of natural numbers (ak) growing to infinity and such that
limk→∞ ak log kk
= 0. Prove that limk→∞ S(kak, k)
k2 = 14 .
Problem 1.66. (Nieuw Archief v. Wiskunde) Let a1 < a2 < a3 < · · · < an < · · · bea sequence of positive integers such that the series
∞∑
i=1
1
ai
converges. Prove that for any i, there exist infinitely many sets of ai consecutiveintegers that are not divisible by aj for all j > i.
Problem 1.67. (Amer. Math. Monthly) Denote by Cn the claim that there exists a setof n consecutive integers such that no two of them are relatively prime. Prove that Cnis true for every n such that 17 ≤ n ≤ 10000.
Problem 1.68. (Schweitzer Competition) Prove that a natural number has more di-visors that can be written in the form 3k + 1, for k ∈ Z, than divisors of the form3m− 1, for m ∈ Z.
Problem 1.69. (Amer. Math. Monthly) A numberN is called deficient if σ(N) < 2Nand abundant if σ(N) > 2N .
1. Let k be fixed. Are there any sequences of k consecutive abundant numbers?2. Show that there are infinitely many 5-tuples of consecutive deficient numbers.
Problem 1.70. (Amer. Math. Monthly) Does there exist a nonconstant polynomialan2 +bn+cwith integer coefficients such that for any natural numbersm all its primefactors pi are congruent to 3 modulo 4? Prove that for any nonconstant polynomialf with integer coefficients and anym ∈ Z there exist a prime number p and a naturalnumber n such that p divides f (n) and p ≡ 1 (mod m).
2
Algebra and Combinatorics
2.1 Algebra
Problem 2.1. Set Sk,p = ∑p−1i=1 i
k , for natural numbers p and k. If p ≥ 3 is primeand 1 < k ≤ p − 2, show that
Sk,p ≡ 0 (mod p).
Problem 2.2. LetP = a0 +· · ·+anxn andQ = b0 +· · ·+bmxm be two polynomialswith m ≤ n. Then, deg gcd(P,Q) ≥ 1 if and only if there exist two polynomials Kand L, such that deg K ≤ m− 1, deg L ≤ n− 1, andK ·P = L ·Q. Prove that thisis equivalent to the vanishing of the following (n+m)× (n+m) determinant:
det
⎛
⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝
a0 a1 a2 · · · am−1 · · · an−1 an 0 0 · · · 00 a0 a1 · · · am−2 · · · an−2 an−1 an 0 · · · 0......
......
......
......
...
0 0 0 · · · a0 · · · an−m−1 an−m an−m+1 · · · an0 0 0 · · · 0 · · · an−m−2 an−m−1 an−m · · · an−1......
......
......
......
0 0 0 · · · 0 0 a0 a1 a2 · · · amb0 b1 · · · bm−1 bm 0 0 0 0 · · · 00 b0 · · · bm−2 bm−1 bm 0 0 0 · · · 0......
......
......
......
...
0 0 · · · b0 b1 · · · · · · · · · 0 · · · 0
⎞
⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠
= 0.
Problem 2.3. Prove that if P,Q ∈ R[x, y] are relatively prime polynomials, then thesystem of equations
P(x, y) = 0,Q(x, y) = 0,
has only finitely many real solutions.
12 2 Algebra and Combinatorics
Problem 2.4. Let a, b, c, d ∈ R[x] be polynomials with real coefficients. Set
p =∫ x
1ac dt, q =
∫ x
1ad dt, r =
∫ x
1bc dt, s =
∫ x
1bd dt.
Prove that ps − qr is divisible by (x − 1)4.
Problem 2.5. 1. Find the minimum number of elements that must be deleted fromthe set {1, . . . , 2005} such that the set of the remaining elements does not contain twoelements together with their product.
2. Does there exist, for any k, an arithmetic progression with k terms in the infinitesequence
1,1
2, . . . ,
1
2005, . . . ,
1
n, . . .?
Problem 2.6. (Amer. Math. Monthly) Consider a set S of n elements and n+1 subsetsM1, . . . ,Mn+1 ⊂ S. Show that there exist r, s ≥ 1 and disjoint sets of indices{i1, . . . , ir} ∩ {j1, . . . , js} = ∅ such that
r⋃
k=1
Mik =s⋃
k=1
Mjk .
Problem 2.7. Let p be a prime number, and A = {a1, . . . , ap−1} ⊂ Z∗+ a set of
integers that are not divisible by p. Define the map f : P(A) → {0, 1, . . . , p − 1}by
f ({ai1 , . . . , aik }) =k∑
p−1
aip (mod p), and f (∅) = 0.
Prove that f is surjective.
Problem 2.8. Consider the function Fr = xr sin rA + yr sin rB + zr sin rC, wherex, y, z ∈ R, A + B + C = kπ , and r ∈ Z+. Prove that if F1(x0, y0, z0) =F2(x0, y0, z0) = 0, then Fr(x0, y0, z0) = 0, for all r ∈ Z+.
Problem 2.9. Let T (z) ∈ Z[z] be a nonzero polynomial with the property that|T (ui)| ≤ 1 for all values ui that are roots of P(z) = zn − 1. Prove that eitherT (z) is divisible by P(z), or else there exists some k ∈ Z+, k ≤ n − 1, such thatT (z) ± zk is divisible by P(z). The same result holds when instead of P(z), weconsider zn + 1.
Problem 2.10. 1. If the map x �→ x3 from a group G to itself is an injective grouphomomorphism, then G is an abelian.
2. If the map x �→ x3 from a groupG to itself is a surjective group homomorphism,then G is an abelian.
3. Find an abelian group with the property that x �→ x4 is an automorphism.4. What can be said for exponents greater than 4?
2.2 Algebraic Combinatorics 13
Problem 2.11. (Amer. Math. Monthly) Let V be a vector space of dimension n > 0over a field of characteristic p �= 0 and let A be an affine map A : V → V . Provethat there exist u ∈ V and 0 ≤ k ≤ np such that Aku = u.
Problem 2.12. (Putnam Competition 1959) Find the cubic equation the zeros of whichare the cubes of the roots of the equation x3 + ax2 + bx + c = 0.
Problem 2.13. (Putnam Competition 1956) Assume that the polynomials P,Q ∈C[x] have the same roots, possibly with different multiplicities. Suppose, moreover,that the same holds true for the pair P + 1 and Q+ 1. Prove that P = Q.
Problem 2.14. (Amer. Math. Monthly) Determine r ∈ Q, for which 1, cos 2πr, sin 2πrare linearly dependent over Q.
Problem 2.15. (Amer. Math. Monthly)
1. Prove that there exist a, b, c ∈ Z, not all zero, such that |a|, |b|, |c| < 106∣∣∣a + b
√2 + c
√3∣∣∣ < 10−11.
2. Prove that if 0 ≤ |a|, |b|, |c| < 106, a, b, c ∈ Z, and at least one of them is
nonzero, then∣∣∣a + b
√2 + c
√3∣∣∣ > 10−21.
Problem 2.16. (Amer. Math. Monthly) Prove that if n > 2, then we do not have anynontrivial solutions for the equation
xn + yn = zn,
where x, y, z are rational functions. Solutions of the form x = af, y = bf, z = cf ,wheref is a rational function anda, b, c are complex numbers satisfyingan+bn = cn,are called trivial.
Problem 2.17. (Kvant) A table is an n× k rectangular grid drawn on the torus, everybox being assigned an element from Z/2Z. We define a transformation acting ontables as follows. We replace all elements of the grid simultaneously, each elementbeing changed into the sum of the numbers previously assigned to its neighboringboxes. Prove that iterating this transformation sufficiently many times, we alwaysobtain the trivial table filled with zeros, no matter what the initial table was, if andonly if n = 2p and k = 2q for some integers p, q. In this case we say that therespective n× k grid is nilpotent.
2.2 Algebraic Combinatorics
Problem 2.18. Let us consider a four-digit number N whose digits are not all equal.We first arrange its digits in increasing order, then in decreasing order, and finally,we subtract the two obtained numbers. Let T (N) denote the positive difference thusobtained. Show that after finitely many iterations of the transformation T , we obtain6174.
14 2 Algebra and Combinatorics
Problem 2.19. Find an example of a sequence of natural numbers 1 ≤ a1 < a2 <
· · · < an < an+1 < · · · with the property that everym ∈ Z+ can be uniquely writtenas m = ai − aj , for i, j ∈ Z+.
Problem 2.20. (Amer. Math. Monthly) Consider the set of 2n integers {±a1,±a2, . . . ,
±an} and m < 2n. Show that we can choose a subset S such that
1. The two numbers ±ai are not both in S;2. The sum of all elements of S is divisible by m.
Problem 2.21. (Proposed for the International Math. Olympiad) Show that for ev-ery natural number n there exist prime numbers p and q such that n divides theirdifference.
Problem 2.22. An even number, 2n, of knights arrive at King Arthur’s court, eachone of them having at most n − 1 enemies. Prove that Merlin the wizard can assignplaces for them at a round table in such a way that every knight is sitting only next tofriends.
Problem 2.23. (Putnam Competition) Let r, s ∈ Z+. Find the number of 4-tuplesof positive integers (a, b, c, d) that satisfy 3r7s = lcm(a, b, c) = lcm(a, b, d) =lcm(a, c, d) = lcm(b, c, d).
Problem 2.24. (Putnam Competition)
1. Let n ∈ Z+ and p be a prime number. Denote byN(n, p) the number of binomialcoefficients Csn that are not divisible by p. Assume that n is written in base p asn = n0 +n1p+· · ·+nmpm, where 0 ≤ nj < p for all j ∈ {0, 1, . . . , m}. Provethat N(n, p) = (n0 + 1)(n1 + 1) · · · (nj + 1).
2. Write k in base p as k = k0 + k1p + · · · + kjpj , with 0 ≤ kj ≤ p − 1, for all
j ∈ {0, 1, . . . , s}. Prove that
Ckn ≡ Ck0n0Ck1n1
· · ·Ckjnj (mod p).
Problem 2.25. (Putnam Competition) Define the sequence Tn by T1 = 2, Tn+1 =T 2n − Tn + 1, for n ≥ 1. Prove that if m �= n, then Tm and Tn are relatively prime,
and further, that∞∑
i=1
1
Ti= 1.
Problem 2.26. Let α, β > 0 and consider the sequences
[α], [2α], . . . , [kα], . . . ; [β], [2β], . . . , [kβ], . . . ,
where the brackets denote the integer part. Prove that these two sequences takentogether enumerate Z+ in an injective manner if and only if
α, β ∈ R \ Q and1
α+ 1
β= 1.
2.2 Algebraic Combinatorics 15
Problem 2.27. We say that the sets S1, S2, . . . , Sm form a complementary system ifthey make a partition of Z+, i.e., every positive integer belongs to a unique set Si . Letm > 1 and α1, . . . , αm ∈ R+. Then the sets
Si = {[nαi], where n ∈ Z+}form a complementary system only if
m = 2, α−11 + α−1
2 = 1, and α1 ∈ R \ Q.
Problem 2.28. Let f : Z+ → Z+ be an increasing function and set
F(n) = f (n)+ n, G(n) = f ∗(n)+ n,
where f ∗(n) = card({x ∈ Z+; 0 ≤ f (x) < n}). Then {F(n); n ∈ Z+} and{G(n); n ∈ Z+} are complementary sequences. Conversely, any two complemen-tary sequences can be obtained this way using some nondecreasing function f .
Problem 2.29. LetM denote the set of bijective functions f : Z+ → Z+. Prove thatthere is no bijective function between M and Z.
Problem 2.30. (Amer. Math. Monthly) Let F ⊂ Z be a finite set of integers satisfyingthe following properties:
1. For any x ∈ F there exist y, z ∈ F such that x = y + z.2. There exists n such that for any natural number 1 ≤ k ≤ n, and any choice ofx1, . . . , xk ∈ F , their sum x1 + · · · + xk is nonzero.
Prove that card(F ) ≥ 2n+ 2.
Problem 2.31. (Amer. Math. Monthly) For a finite graph G we denote by Z(G) theminimal number of colors needed to color all its vertices such that adjacent verticeshave different colors. This is also called the chromatic number of G.
Prove that the inequality
Z(G) ≥ p2
p2 − 2q
holds if G has p vertices and q edges.
Problem 2.32. LetDk be a collection of subsets of the set {1, . . . , n} with the propertythat whenever A �= B ∈ Dk , then card(A ∩ B) ≤ k, where 0 ≤ k ≤ n − 1. Provethat
card(Dk) ≤ C0n + C1
n + C2n + · · · + Ck+1
n .
Problem 2.33. Prove that
1
p!
n∑
k=0
(−1)n−kCknkp =
⎧⎪⎪⎪⎪⎪⎪⎪⎨
⎪⎪⎪⎪⎪⎪⎪⎩
0, if 0 ≤ p < n,
1, if p = n,
n/2, if p = n+ 1,n(3n+1)
24 , if p = n+ 2,n2(n+1)
48 , if p = n+ 3,n(15n3+30n2+5n+1)
1152 , if p = n+ 4.
16 2 Algebra and Combinatorics
Problem 2.34. Write lcm(a1, . . . , an) in terms of the various gcd(ai, . . . , aj ) forsubsets of {a1, . . . , an}.Problem 2.35. Let f (n) be the number of ways in which a convex polygon with n+1sides can be divided into regions delimited by several diagonals that do not intersect(except possibly at their endpoints). We consider as distinct the dissection in whichwe first cut the diagonal a and next the diagonal b from the dissection in which wefirst cut the diagonal b and next the diagonal a. It is easy to compute the first valuesof f (n), as follows: f (1) = 1, f (2) = 1, f (3) = 3, f (4) = 11, f (5) = 45. Find thegenerating function F(x) = ∑
f (n)xn and an asymptotic formula for f (n).
Problem 2.36. Find the permutation σ : (1, . . . , n) → (1, . . . , n) such that
S(σ) =n∑
i=1
|σ(i)− i|
is maximal.
Problem 2.37. (Amer. Math. Monthly) On the set Sn of permutations of {1, . . . , n}we define an invariant distance function by means of the formula
d(σ, τ ) =n∑
i=1
|σ(i)− τ(i)|.
What are the values that d could possibly take?
Problem 2.38. (Preliminaries for the International Math. Olympiad, Romania) Theset M = {1, 2, . . . , 2n} is partitioned into k sets M1, . . . ,Mk , where n ≥ k3 + k.Show that there exist i, j ∈ {1, . . . , k} for which we can find k + 1 distinct evennumbers 2r1, . . . , 2rk+1 ∈ Mi with the property that 2r1 − 1, . . . , 2rk+1 − 1 ∈ Mj .
Problem 2.39. (Preliminaries for the International Math. Olympiad, Great Britain)Let S be the set of odd integers not divisible by 5 and smaller than 30m, wherem ∈ Z
∗+. Find the smallest k such that every subset A ⊂ S of k elements containstwo distinct integers, one of which divides the other.
Problem 2.40. Prove that∏
1≤j<i≤nai−aji−j is a natural number whenever a1 ≤ a2 ≤
· · · ≤ an are integers.
Problem 2.41. Is there an infinite setA ⊂ Z+ such that for all x, y ∈ A neither x norx + y is a perfect power, i.e., ak , for k ≥ 2? More generally, is there an infinite setA ⊂ Z+ such that for any nonempty finite collection xi ∈ A, i ∈ J , the sum
∑i∈J xi
is not a perfect power?
Problem 2.42. (Amer. Math. Monthly) Let
f (n) = maxAAA
...
Ak
32
1 ,
where n = A1 +· · ·+Ak . Thus, f (1) = 1, f (2) = 2, f (3) = 3, f (4) = 4, f (5) =9, f (6) = 27, f (7) = 512, etc. Determine f (n).
2.2 Algebraic Combinatorics 17
Problem 2.43. Consider a setM withm elements andA1, . . . , An distinct subsets ofM such that card(Ai ∩ Aj) = r ≥ 1 for all 1 ≤ i �= j ≤ n. Prove that n ≤ m.
Problem 2.44. (Amer. Math. Monthly) Setπ(n) for the number of prime numbers lessthan or equal to n. Prove that there are at most π(n) numbers 1 < a1 < · · · < ak ≤ n
with gcd(ai, aj ) = 1.
Problem 2.45. (Nieuw Archief v. Wiskunde) Prove that for every k, there exists n suchthat the nth term of the Fibonacci sequence Fn is divisible by k. Recall that Fn isdetermined by the recurrence: Fn+2 = Fn + Fn+1, for n ≥ 0, where the first termsare F0 = 0, F1 = 1.
Problem 2.46. (Amer. Math. Monthly) Consider a set of consecutive integers C +1, C + 2, . . . , C + n, where C > nn−1. Show that there exist distinct prime numbersp1, p2, . . . , pn such that C + j is divisible by pj .
Problem 2.47. (Nieuw Archief v. Wiskunde) Let p be a prime number and f (p) thesmallest integer for which there exists a partition of the set {2, 3, . . . , p} into f (p)classes such that whenever a1, . . . , ak belong to the same class of the partition, theequation
k∑
i=1
xiai = p
does not have solutions in nonnegative integers. Estimate f (p).
Problem 2.48. (Schweitzer Competition) Consider m distinct natural numbers aismaller than N such that lcm(ai, aj ) ≤ N for all i, j . Prove that m ≤ 2
[√N].
Problem 2.49. The setM ⊂ Z+ is calledA-sum-free, whereA = (a1, a2, . . . , ak) ∈Zk+, if for any choice of x1, x2, . . . , xk ∈ M we have a1x1 + a2x2 + · · · + akxk �∈ M .
IfA,B are two vectors, we define f (n;A,B) as the greatest number h such that thereexists a partition of the set of consecutive integers {n, n + 1, . . . , h} into S1 and S2such that S1 is A-sum-free and S2 is B-sum-free. Assume that B = (b1, b2, . . . , bm)
and that the conditions below are satisfied:
a1 + a2 + · · · + ak = b1 + b2 + · · · + bm = s,
andmin
1≤j≤k aj = min1≤j≤m bj = 1, k,m ≥ 2.
Prove that f (n;A,B) = ns2 + n(s − 1)− 1.
Problem 2.50. (Amer. Math. Monthly) Let 1 ≤ a1 < a2 < · · · < an < 2n be asequence of natural numbers for n ≥ 6. Prove that
mini,j
lcm(ai, aj ) ≤ 6([n
2
]+ 1
).
Moreover, the constant 6 is sharp.
18 2 Algebra and Combinatorics
Problem 2.51. (Amer. Math. Monthly) Let 1 ≤ a1 < a2 < · · · < ak < n be such thatgcd(ai, aj ) �= 1 for all 1 ≤ i < j ≤ k. Determine the maximum value of k.
Problem 2.52. (International Math. Olympiad 1978) Consider the increasing se-quence f (n) ∈ Z+, 0 < f (1) < f (2) < · · · < f (n) < · · · . It is known thatthe nth element in increasing order among the positive integers that are not terms ofthis sequence is f (f (n))+ 1. Find the value of f (240).
Problem 2.53. (Amer. Math. Monthly) We define inductively three sequences of in-tegers (an), (bn), (cn) as follows:
1. a1 = 1, b1 = 2, c1 = 4;2. an is the smallest integer that does not belong to the set
{a1, . . . , an−1, b1, . . . , bn−1, c1, . . . , cn−1};3. bn is the smallest integer that does not belong to the set
{a1, . . . , an−1, an, b1, . . . , bn−1, c1, . . . , cn−1};4. cn = 2bn + n− an.
Prove that0 < n
(1 + √
3)
− bn < 2 for all n ∈ Z+.
2.3 Geometric Combinatorics
Problem 2.54. We consider n points in the plane that determine C2n segments, and to
each segment one associates either +1 or −1. A triangle whose vertices are amongthese points will be called negative if the product of numbers associated to its sidesis negative. Show that if n is even, then the number of negative triangles is even.Moreover, for odd n, the number of negative triangles has the same parity as thenumber p of segments labeled −1.
Problem 2.55. (Amer. Math. Monthly) Given n find a finite set S consisting of naturalnumbers larger than n with the property that for any k ≥ n the k × k square can betiled by a family of si × si squares, where si ∈ S.
Problem 2.56. We consider 3n points A1, . . . , A3n in the plane whose positions aredefined recursively by means of the following rule: first, the triangle A1A2A3 isequilateral; further, the points A3k+1, A3k+2, and A3k+3 are the midpoints of thesides of the triangle A3kA3k−1A3k−2. Let us assume that the 3n points are coloredwith two colors. Show that for n ≥ 7 there exists at least one isosceles trapezoidhaving vertices of the same color.
Problem 2.57. Is there a coloring of all lattice points in the plane using only twocolors such that there are no rectangles with all vertices of the same color whose sideratio belongs to
{1, 1
2 ,13 ,
23
}?
2.3 Geometric Combinatorics 19
Problem 2.58. (Amer. Math. Monthly) Let G be a planar graph and let P be a pathin G. We say that P has a (transversal) self-intersection in the vertex v if the pathhas a (transversal) self-intersection from the curve-theoretic viewpoint. Let us give anexample: Take the point 0 in the plane and the segments 01, 02, 03, 04 going counter-clockwise around 0. Then a path traversing first 103 and then 204 has a (transversal)self-intersection at 0, while a path going first along 102 and further on 304 does nothave a (transversal) self-intersection.
Prove that any connected planar graph G with only even-degree vertices admitsan Eulerian circuit without self-intersections. Recall that an Eulerian circuit is a pathalong the edges of the graph that passes precisely once along each edge of the graph.
Problem 2.59. (Hungarian Competition) Let us consider finitely many points in theplane that are not all collinear. Assume that one associates to each point a numberfrom the set {−1, 0, 1} such that the following property holds: for any line determinedby two points from the set, the sum of numbers associated to all points lying on thatline equals zero. Show that, if the number of points is at least three, then to each pointone is associated with 0.
Problem 2.60. (Amer. Math. Monthly) If one has a set of squares with total areasmaller than 1, then one can arrange them inside a square of side length
√2, without
any overlaps.
Problem 2.61. Prove that for each k there exist k points in the plane, no three collinearand having integral distance from each other. If we have an infinite set of points withintegral distances from each other, then all points are collinear.
Problem 2.62. (International Math. Olympiad 1984) Let O,A be distinct points inthe plane. For each point x in the plane, we write α(x) = xOA (counterclockwise).Let C(x) be the circle of center O and radius |Ox| + α(x)
|Ox| . If the points in the planeare colored with finitely many colors, then there exists a point y with α(y) > 0 suchthat the color of y also belongs to the circle C(y).
Problem 2.63. (Amer. Math. Monthly) Let k, n ∈ Z+.
1. Assume that n−1 ≤ k ≤ n(n−1)2 . Show that there exist n distinct points x1, ..., xn
on a line that determine exactly k distinct distances |xi − xj |.2. Suppose that [n2 ] ≤ k ≤ n(n−1)
2 . Then there exist n points in the plane thatdetermine exactly k distinct distances.
3. Prove that for any ε > 0, there exists some constant n0 = n0(ε) such that for anyn > n0 and εn < k <
n(n−1)2 , there exist n points in the plane that determine
exactly k distinct distances.
Problem 2.64. (International Math. Olympiad 1978) Show that it is possible to pack2n(2n+ 1) nonoverlapping pieces having the form of a parallelepiped of dimensions1 × 2 × (n+ 1) in a cubic box of side 2n+ 1 if and only if n is even or n = 1.
20 2 Algebra and Combinatorics
Problem 2.65. (Putnam Competition 1964) Let F be a finite subset of R with theproperty that any value of the distance between two points from F (except for thelargest one) is attained at least twice, i.e., for two distinct pairs of points. Prove thatthe ratio of any two distances between points of F is a rational number.
3
Geometry
3.1 Synthetic Geometry
Problem 3.1. Let I be the center of the circle inscribed in the triangle ABC andconsider the points α, β, γ situated on the perpendiculars from I on the sides of thetriangle ABC such that
|Iα| = |Iβ| = |Iγ |.Prove that the lines Aα,Bβ,Cγ are concurrent.
Problem 3.2. We consider the angle xOy and a pointA ∈ Ox. Let (C) be an arbitrarycircle that is tangent to Ox and Oy at the points H and D, respectively. Set AE forthe tangent line drawn from A to the circle (C) that is different from AH . Show thatthe lineDE passes through a fixed point that is independent of the circle (C) chosenabove.
Problem 3.3. Let C be a circle of center O and A a fixed point in the plane. For anypoint P ∈ C, let M denote the intersection of the bisector of the angle AOP withthe circle circumscribed about the triangle AOP . Find the geometric locus of M asP runs over the circle C.
Problem 3.4. Let ABC be an isosceles triangle having |AB| = |AC|. If AS is aninterior Cevian that intersects the circle circumscribed aboutABC at S, then describethe geometric locus of the center of the circle circumscribed about the triangle BST ,where {T } = AS ∩ BC.
Problem 3.5. LetAB,CD,EF be three chords of length one on the unit circle. Thenthe midpoints of the segments |BC|, |DE|, and |AF | form an equilateral triangle.
Problem 3.6. (Amer. Math. Monthly) Denote by P the set of points of the plane. Let� : P ×P → P be the following binary operation:A�B = C, whereC is the uniquepoint in the plane such thatABC is an oriented equilateral triangle whose orientationis counterclockwise. Show that � is a nonassociative and noncommutative operationsatisfying the following “medial property”:
22 3 Geometry
(A � B) � (C � D) = (A � C) � (B � D).
Problem 3.7. Consider two distinct circles C1 and C2 with nonempty intersectionand let A be a point of intersection. Let P,R ∈ C1 and Q,S ∈ C2 be such thatPQ and RS are the two common tangents. Let U and V denote the midpoints of thechords PR and QS. Prove that the triangle AUV is isosceles.
Problem 3.8. (Amer. Math. Monthly) If the planar triangles AUV, VBU , and UVCare directly similar to a given triangle, then so is ABC. Recall that two triangles aredirectly similar if one can obtain one from the other using a homothety with positiveratio, rotations and translations.
Problem 3.9. (Amer. Math. Monthly) Find, using a straightedge and a compass, thedirectrix and the focus of a parabola. Recall that the parabola is the geometric locusof those points P in the plane that are at equal distance from a point O (called thefocus) and a line d called the directrix.
Problem 3.10. Prove that if M is a point in the interior of a circle and AB ⊥ CD
are two chords perpendicular at M , then it is possible to construct an inscribablequadrilateral with the following lengths:
∣∣|AM| − |MB|∣∣, |AM| + |MB|, ∣∣|DM| − |MC|∣∣, |DM| + |MC|.Problem 3.11. (Amer. Math. Monthly) If the Euler line of a triangle passes throughthe Fermat point, then the triangle is isosceles.
Problem 3.12. Consider a point M in the interior of the triangle ABC, and chooseA′ ∈ AM,B ′ ∈ BM , and C′ ∈ CM . Let P,Q,R, S, T , and U be the intersectionsof the sides of ABC and A′B ′C′. Show that PS, TQ, and RU meet at M .
Problem 3.13. Show that if an altitude in a tetrahedron crosses two other altitudes,then all four altitudes are concurrent.
Problem 3.14. Three concurrent Cevians in the interior of the triangle ABC meetthe corresponding opposite sides atA1, B1, C1. Show that their common intersectionpoint is uniquely determined if |BA1|, |CB1|, and |AC1| are equal.
Problem 3.15. (International Math. Olympiad 1983) LetABCD be a convex quadri-lateral with the property that the circle of diameterAB is tangent to the lineCD. Provethat the circle of diameter CD is tangent to the line AB if and only if AD is parallelto BC.
Problem 3.16. Let A′, B ′, C′ be points on the sides BC,CA,AB of the triangleABC. LetM1,M2 be the intersections of the circleA′B ′C′ with the circleABA′ andlet N1, N2 be the analogous intersections of the circle A′B ′C′ with the circle ABB ′.
1. Prove thatM1M2, N1N2, AB are either parallel or concurrent, in a point that wedenote by A′
1;
3.2 Combinatorial Geometry 23
2. Prove that the analogously defined points A′1, B
′1, C
′1 are collinear.
Problem 3.17. (Putnam Competition) A circumscribable quadrilateral of area S =√abcd is inscribable.
Problem 3.18. (Nieuw Archief v. Wiskunde) Let O be the center of the circumcircle,Ge the Gergonne point, Na the Nagel point, and G1, N1, the isogonal conjugates ofG and N , respectively. Prove thatG1, N1, andO are collinear (see also the Glossaryfor definitions of the important points in a triangle).
3.2 Combinatorial Geometry
Problem 3.19. Consider a rectangular sheet of paper. Prove that given any ε > 0,one can use finitely many foldings of the paper along its sides in either 2 equal partsor 3 equal parts to obtain a rectangle whose sides are in ratio r for some r satisfying1 − ε ≤ r ≤ 1 + ε.
Problem 3.20. (Amer. Math. Monthly) Show that there exist at most three points onthe unit circle with the distance between any two being greater than
√2.
Problem 3.21. (Komal) A convex polygon with 2n sides has at least n diagonals notparallel to any of its sides.
Problem 3.22. Let d be the sum of the lengths of the diagonals of a convex polygonP1 . . . Pn and p its perimeter. Prove that for n ≥ 4, we have
n− 3 < 2d
p<
[n2
] [n+ 1
2
]− 2.
Problem 3.23. (Amer. Math. Monthly) Find the convex polygons with the propertythat the function D(p), which is the sum of the distances from an interior point p tothe sides of the polygon, does not depend on p.
Problem 3.24. Prove that a sphere of diameter 1 cannot be covered by n strips ofwidth li if
∑ni=1 li < 1. Prove that a circle of diameter 1 cannot be covered by n strips
of width li if∑ni=1 li < 1.
Problem 3.25. (Kvant) Consider n points lying on the unit sphere. Prove that thesum of the squares of the lengths of all segments determined by the n points is lessthan n2.
Problem 3.26. (Kvant) The sum of the vectors−−→OA1, . . . ,
−−→OAn is zero, and the sum
of their lengths is d . Prove that the perimeter of the polygon A1 . . . An is greater than4d/n.
Problem 3.27. (Amer. Math. Monthly) Find the largest numbers ak , for 1 ≤ k ≤ 7,with the property that for any point P lying in the unit cube with vertices A1 . . . A8,at least k among the distances |PAj | to the vertices are greater or equal than ak .
24 3 Geometry
Problem 3.28. (Amer. Math. Monthly) The line determined by two points is said tobe admissible if its slope is equal to 0, 1,−1, or ∞. What is the maximum numberof admissible lines determined by n points in the plane?
Problem 3.29. If A = {z1, . . . , zn} ⊂ C, then there exists a subset B ⊂ A such that
∣∣∑
z∈Bz∣∣ ≥ π−1
n∑
i=1
|zi |.
Problem 3.30. (Amer. Math. Monthly) Let A1, . . . , An be the vertices of a regularn-gon inscribed in a circle of center O. Let B be a point on the arc of circle A1An
and set θ = AnOB. If we set ak = |BAk|, then find the sum
n∑
k=1
(−1)kak
in terms of θ .
Problem 3.31. (Amer. Math. Monthly) Consider n distinct complex numbers zi ∈ C
such thatmini �=j |zi − zj | ≥ max
i≤n |zi |.What is the greatest possible value of n?
Problem 3.32. The interior of a triangle can be tiled by n ≥ 9 pentagonal convexsurfaces.What is the minimal value ofn such that a triangle can be tiled byn hexagonalstrictly convex surfaces?
Problem 3.33. (Putnam Competition) We say that a transformation of the plane is acongruence if it preserves the length of segments. Two subsets are congruent if thereexists a congruence sending one subset onto the other. Show that the unit disk cannotbe partitioned into two congruent subsets.
Problem 3.34. Prove that the unit disk cannot be partitioned into two subsets ofdiameter strictly smaller than 1, where the diameter of a set is the supremum distancebetween two of its points.
Problem 3.35. A continuous planar curve L has extremities A and B at distance|AB| = 1. Show that for any natural number n there exists a chord determined bytwo points C,D ∈ L that is parallel to AB and whose length |CD| equals 1
n.
Problem 3.36. The diameter of a set is the supremum of the distance between twoof its points. Prove that any planar set of unit diameter can pe partitioned into three
parts of diameter no more than√
32 .
Problem 3.37. 1. Prove that a finite set of n points in R3 of unit diameter can be
covered by a cube of side length 1 − 23n(n−1) .
3.2 Combinatorial Geometry 25
2. Prove that any planar set of n points having unit diameter can pe partitioned into
three parts of diameter less than√
32 cos 2π
3n(n−1) .
Problem 3.38. Prove that any convex set in Rn of unit diameter having a smooth
boundary can be partitioned into n+ 1 parts of diameter d < 1.
Problem 3.39. Let D be a convex body in R3 and let σ(D) = supπ area(π ∩ D),
where the supremum is taken over all positions of the variable plane π . Prove thatDcan be divided into two parts D1 and D2 such that σ(Di) < σ(D).
Problem 3.40. If we have k vectors v1, v2, . . . , vk in Rn and k ≤ n + 1, then there
exist two vectors making an angle θ with cos θ ≥ − 1k−1 . Equality holds only when
the endpoints of the vectors form a regular (k − 1)-simplex.
Problem 3.41. 1. Consider a finite family of bounded closed convex sets in theplane such that any three members of the family have nonempty intersection.Prove that the intersection of all members of the family is nonempty.
2. A set of unit diameter in R2 can be covered by a ball of radius
√13 .
Problem 3.42. (Amer. Math. Monthly) Let T be a right isosceles triangle. Find thedisk D such that the difference between the areas of T ∪D and T ∩D is minimal.
Problem 3.43. (Amer. Math. Monthly) Let r be the radius of the incircle of an arbitrary
triangle lying in the closed unit square. Prove that r ≤√
5−14 .
Problem 3.44. Let P be a point in the interior of the tetrahedron ABCD, with theproperty that |PA| + |PB| + |PC| + |PD| is minimal. Prove that APB = CPD
and that these angles have a common bisector.
Problem 3.45. (Putnam Competition 1948) Let OA1, . . . , OAn be n linearly inde-pendent vectors of lengths a1, . . . , an. We construct the parallelepipedH having thesevectors as sides. Then consider the n altitudes inH as a new set of vectors and further,construct the parallelepiped E associated with the altitudes. If h is the volume of Hand e the volume of E, then prove that
he = (a1 . . . an)2.
Problem 3.46. Let F be a symmetric convex body in R3 and let AF,λ denote the
family of all sets homothetic to F in the ratio λ that have only boundary points incommon with F . Set hF (λ) for the greatest integer k such that AF,λ contains k setswith pairwise disjoint interiors. Prove that
hF (λ) ≤ (1 + 2λ)3 − 1
λ3 .
26 3 Geometry
Problem 3.47. Let � denote the square of equations |xi | ≤ 1, i = 1, 2, in the plane,and let A = (a1|a2) be an arbitrary nonsingular 2 × 2 matrix partitioned into twocolumns. We identify each column with a vector in R
2. Prove that the followinginequality holds:
minA
maxx∈�
∣∣∣∣〈a1, x〉〈a2, x〉
det A
∣∣∣∣ = 1
2,
where 〈x, y〉 = x1x2 + y1y2 is the usual scalar product.
Problem 3.48. We denote by δ(r) the minimal distance between a lattice point andthe circle C(O, r) of radius r centered at the originO of the coordinate system in theplane. Prove that
limr→∞ δ(r) = 0.
Problem 3.49. (Putnam Competition) Consider a curve C of length l that divides thesurface of the unit sphere into two parts of equal area. Show that l ≥ 2π .
Problem 3.50. (Amer. Math. Monthly) Let K be a planar closed curve of length 2π .Prove that K can be inscribed in a rectangle of area 4.
Problem 3.51. 1. Consider a family of plane convex sets with area a, perimeter p,and diameter d. If the family covers area A, then there exists a subfamily withpairwise disjoint interiors that covers at least area λA, where λ = a
a+pd+πd2 .2. Assume that any two members of the family have nonempty intersection. Prove
that there exists then a subfamily with pairwise disjoint interiors that covers areaat least μA, where μ = a
πd2 .
Problem 3.52. (Putnam Compettion 1957) Let C be a regular polygon with k sides.Prove that for every n there exists a planar set S(n) ⊂ R
2 such that any subsetconsisting of n points of S(n) can be covered by C, but S(n) itself cannot be includedin C.
Problem 3.53. Let M be a convex polygon and let S1, . . . , Sn be pairwise disjointdisks situated in the interior of M . Does there exist a partition M = D1 ∪ · · · ∪Dnsuch thatDi are convex disjoint polygons, each of which contains precisely one disk?
Problem 3.54. Consider an inscribable n-gon partitioned by means of n − 2 nonin-tersecting diagonals into n− 2 triangles. Prove that the sum of the radii of the circlesinscribed in these triangles does not depend on the particular partition.
Problem 3.55. (Elemente der Mathematik) Prove that in an ellipse having axes oflengths a and b and total length L, we have L > π(a + b).
Problem 3.56. (Kvant) Let F be a convex planar domain and F ′ denote its imageby a homothety of ratio − 1
2 . Is it true that one can translate F ′ in order for it to becontained in F ? Can the constant 1
2 be improved? Generalize to n dimensions.
3.3 Geometric Inequalities 27
Problem 3.57. (Nieuw Archief v.Wiskunde)A classical theorem, due to Cauchy, statesthat a strictly convex polyhedron in R
3 whose faces are rigid must be globally rigid.Here, rigidity means continuous rigidity in the sense that any continuous deformationof the polyhedron in R
3 that keeps the lengths of edges fixed is the restriction of adeformation of rigid Euclidean motions of three-space. Prove that a 3-dimensionalcube immersed in R
n remains rigid for all n > 3.
Problem 3.58. Consider finitely many great circles on a sphere such that not all ofthem pass through the same point. Show that there exists a point situated on exactlytwo circles. Deduce that if we have a set of n points in the plane, not all of them lyingon the same line, then there must exist one line passing through precisely two pointsof the given set.
Problem 3.59. Given a finite set of points in the plane labeled with +1 or −1, andnot all of them collinear, show that there exists a line determined by two points in theset such that all points of the set lying on that line are of the same sign.
Problem 3.60. (Amer. Math. Monthly) If Q is a given rectangle and ε > 0, then Qcan be covered by the union of a finite collection S of rectangles with sides parallelto those of Q in such a way that the union of every nonoverlapping subcollection ofS has area less than ε.
Problem 3.61. Prove that the 3-dimensional ball cannot be partitioned into three setsof strictly smaller diameter.
3.3 Geometric Inequalities
Problem 3.62. If a, b, c, r, R are the usual notations in the triangle, show that
1
2rR≤ 1
3
(∑ 1
a
)2
≤∑ 1
a2 ≤ 1
4r2 .
Problem 3.63. (Amer. Math. Monthly) If a, b, c are the sides of a triangle, then prove
that (b+c)2
4bc ≤ mawa
and b2+c2
2bc ≤ maka
, where ma,wa, ka denote respectively the lengthsof the median, bisector, and altitude issued from A.
Problem 3.64. (Putnam Competition) If S(x, y, z) is the area of a triangle with sidesx, y, z, prove that
√S(a, b, c)+ √
S(a′, b′, c′) ≤ √S(a + a′, b + b′, c + c′).
Problem 3.65. (Nieuw Archief v. Wiskunde) It is known that in any triangle we havethe inequality
3√
3r ≤ p ≤ 2R + (3√
3 − 4)r,
where p denotes the semiperimeter. Prove that in an obtuse triangle we have
(3 + 2√
2)r < p < 2R + r.
28 3 Geometry
Problem 3.66. Prove the Euler inequality
R ≥ 2r.
Problem 3.67. Prove that in a triangle we have the inequalities
36r2 ≤ a2 + b2 + c2 ≤ 9R2.
Problem 3.68. (Amer. Math. Monthly)
1. Let ABC and A′B ′C′ be two triangles. Prove that
a2
a′ + b2
b′ + c2
c′≤ R2 (a
′ + b′ + c′)2
a′b′c′.
2. Derive that
a2 + b2 + c2 ≤ 9R2,
cosA cosB cosC ≤ 1
8.
Problem 3.69. (Elemente der Mathematik) Prove that the following inequalities holdin a triangle:
4∑
cyclic
hAhB ≤ 12S√
3 ≤ 54Rr ≤ 3∑
cyclic
ab ≤ 4∑
cyclic
rArB.
Problem 3.70. (Amer. Math. Monthly) Prove that in an any triangle ABC, we have√
1 + 8 cos2 B
sinA+
√1 + 8 cos2 C
sinB+
√1 + 8 cos2 A
sinC≥ 6.
Problem 3.71. Let P be a point in the interior of the triangle ABC. We denote byRa,Rb, Rc the distances from P toA,B,C and by ra, rb, rc the distances to the sidesBC,CA,AB. Prove that
∑
cyclic
R2a sin2 A ≤ 3
∑
cyclic
r2a ,
with equality if and only if P is the Lemoine point (i.e., the symmedian point).
Problem 3.72. Prove the inequalities
16Rr − 5r2 ≤ p2 ≤ 4R2 + 4Rr + 3r2.
Problem 3.73. Prove the following inequalities, due to Roché:
2R2 + 10Rr − r2 − 2(R − 2r)√R2 − 2Rr
≤ p2 ≤ 2R2 + 10Rr − r2 + 2(R − 2r)√R2 − 2Rr.
4
Analysis
Problem 4.1. (Amer. Math. Monthly) Prove that z ∈ C satisfies |z| − �z ≤ 12 if and
only if z = ac, where |c − a| ≤ 1. We denote by �z the real part of the complexnumber z.
Problem 4.2. Let a, b, c ∈ R be such that a + 2b + 3c ≥ 14. Prove thata2 + b2 + c2 ≥ 14.
Problem 4.3. Let fn(x) denote the Fibonacci polynomial, which is defined by
f1 = 1, f2 = x, fn = xfn−1 + fn−2.
Prove that the inequality
f 2n ≤ (x2 + 1)2(x2 + 2)n−3
holds for every real x and n ≥ 3.
Problem 4.4. (Amer. Math. Monthly) Prove the inequality
min((b − c)2, (c − a)2, (a − b)2
)≤ 1
2
(a2 + b2 + c2
).
Generalize to min1≤k<i≤n(ak − ai)2.
Problem 4.5. Let a1, . . . , an be the lengths of the sides of a polygon andP its perime-ter. Then
a1
P − a1+ a2
P − a2+ · · · + an
P − an≥ n
n− 1.
Problem 4.6. Assume that a, b, and c are positive numbers that cannot be the sidesof a triangle. Then the following inequality holds:
(abc)2(a + b + c)2(a + b − c)(a − b + c)(b − a + c)
≥(a2 + b2 + c2
)3 (a2 + b2 − c2
) (a2 − b2 + c2)(b2 − a2 + c2
).
30 4 Analysis
Problem 4.7. (Amer. Math. Monthly) Prove that sin2 x < sin x2, for 0 < x ≤ √π2 .
Problem 4.8. Let P(x) = a0 + a1x + · · · + anxn be a real polynomial of degree
n ≥ 2 such that
0 < a0 < −[n2
]∑
j=1
1
2k + 1a2k.
Show that P(x) has a real zero in (−1, 1).
Problem 4.9. (Amer. Math. Monthly) Find the minimum of β and the maximum of αfor which (
1 + 1
n
)n+α≤ e ≤
(1 + 1
n
)n+β
holds for all n ∈ Z+.
Problem 4.10. (International Math. Olympiad 1984) Prove that for nonnegative x, y,and z such that x + y + z = 1, the following inequality holds:
0 ≤ xy + yz+ zx − 2xyz ≤ 7
27.
Problem 4.11. (Putnam Competition 1948) Consider the sequence of nonzero com-plex numbers a1, . . . , an, . . . with the property that |ar − as | > 1 for r �= s. Provethat ∞∑
n=1
1
a3n
converges.
Problem 4.12. (Amer. Math. Monthly) Consider the sequence Sn given by
Sn = n+ 1
2n+1
n∑
i=1
2i
i.
Find limn→∞ Sn.
Problem 4.13. (Putnam Competition 1951) Prove that whenever a, b > 0 we have∫ 1
0
ta−1
1 + tbdt = 1
a− 1
a + b+ 1
a + 2b− 1
a + 3b+ · · · .
Problem 4.14. (Putnam Competition 1951) Let a, b, c, d ∈ Z∗+, and r = 1 − a
b− c
d.
If r > 0 and a + c ≤ 1982, then r > 119833 .
Problem 4.15. (Amer. Math. Monthly) Let ai ∈ R. Prove that
nmin(ai) ≤n∑
i=1
ai − S ≤n∑
i=1
ai + S ≤ nmax(ai),
where (n− 1)S2 = ∑1≤i<j≤n(ai − aj )
2, with S �= 0, and with equality if and onlyif a1 = · · · = an.
4 Analysis 31
Problem 4.16. (Schweitzer Competition) Consider a periodic function f : R → R,of period 1 that is nonnegative, concave on (0, 1) and continuous at 0. Prove that forall real numbers x and natural numbers n the following inequality f (nx) ≤ nf (x) issatisfied.
Problem 4.17. (Amer. Math. Monthly) Let (an) be a sequence of positive numberssuch that
limn→∞
(a1 + · · · + an)
n< ∞, lim
n→∞an
n= 0.
Does this imply that limn→∞a2
1+···+a2n
n2 = 0?
Problem 4.18. (Amer. Math. Monthly) Show that for a fixed m ≥ 2, the followingseries converges for a single value of x:
S(x) = 1 + ...+ 1
m− 1− x
m+ 1
m+ 1+ · · · + 1
2m− 1− x
2m
+ 1
2m+ 1+ · · · + 1
3m− 1− x
3m+ · · · .
Problem 4.19. (Amer. Math. Monthly) Prove that if zi ∈ C, 0 < |zi | ≤ 1, then|z1 − z2| ≤ | log z1 − log z2|.Problem 4.20. (Nieuw Archief v. Wiskunde) The roots of the function of a complexvariable
ζ4(s) = 1 + 2s + 3s + 4s
are simple.
Problem 4.21. (Putnam Competition 1953) Compute
I =∫ π/2
0log sin x dx.
Problem 4.22. (Putnam Competition 1957) Let f a be positive, monotonically de-creasing function on [0, 1]. Prove the following inequality:
∫ 10 xf
2(x)dx∫ 1
0 xf (x)dx≤
∫ 10 f
2(x)dx∫ 1
0 f (x)dx.
Problem 4.23. (Schweitzer Competition) Prove that every real number x such that0 < x ≤ 1 can be represented as an infinite sum
x =∞∑
k=1
1
nk,
where the nk are natural numbers such that nk+1nk
∈ {2, 3, 4}.
32 4 Analysis
Problem 4.24. (Amer. Math. Monthly) Let Sn denote the set of polynomials p(z) =zn + an−1z
n−1 + · · · + a1z+ 1, with ai ∈ C. Find
Mn = minp∈Sn
(max|z|=1
|p(z)|).
Problem 4.25. (Amer. Math. Monthly) Find an explicit formula for the value of Fnthat is given by the recurrence Fn = (n+2)Fn−1 −(n−1)Fn−2 and initial conditionsF1 = a, F2 = b.
Problem 4.26. Find the positive functions f (x, y) and g(x, y) satisfying the follow-ing inequalities:
(
n∑
i=1
aibi)2 ≤ (
n∑
i=1
f (ai, bi))(
n∑
i=1
g(ai, bi)) ≤ (
n∑
i=1
a2i )(
n∑
i=1
b2i )
for all ai, bi ∈ R and n ∈ Z+.
Problem 4.27. (Putnam Competition 1946) Let g ∈ C1(R) be a smooth function suchthat g(0) = 0 and |g′(x)| ≤ |g(x)|. Prove that g(x) = 0.
Problem 4.28. (Putnam Competition 1947) Let a1, b1, c1 ∈ R+ such that a1 + b1 +c1 = 1 and define
an+1 = a2n + 2bncn, bn+1 = b2
n + 2ancn, cn+1 = c2n + 2anbn.
Prove that the sequences (an), (bn), (cn) have the same limit and find that limit.
Problem 4.29. (Kvant) Consider the sequence (an), given by a1 = 0, a2n+1 = a2n =n− an. Prove that an = n
3 , for infinitely many values of n. Does there exist an n suchthat |an − n
3 | > 2005? Also, prove that
limn→
an
n= 1
3.
Problem 4.30. Compute the integral
f (a) =∫ 1
0
log(x2 − 2x cos a + 1)
xdx.
Problem 4.31. (Amer. Math. Monthly) Let −1 < a0 < 1, and define an =( 1
2 (1 + an−1))1/2
for n ≥ 1. Find the limits A,B, and C of the sequences
An = 4n(1 − an), Bn = a1 · · · an, Cn = 4n(B − a1a2 · · · an).Problem 4.32. (Amer. Math. Monthly) Let consider the sequence given by the recur-rence
a1 = a, an = a2n−1 − 2.
Determine those a ∈ R for which (an) is convergent.
4 Analysis 33
Problem 4.33. Let 0 < a < 1 and I = (0, a). Find all functions f : I → R
satisfying at least one of the conditions below:
1. f is continuous and f (xy) = xf (y)+ yf (x).2. f (xy) = xf (x)+ yf (y).
Problem 4.34. If a, b, c, d ∈ C, ac �= 0, prove that
max(|ac|, |ad + bc|, |bd|)max(|a|, |b|)max(|c|, |d|) ≥ −1 + √
5
2.
Problem 4.35. (Putnam Competition) Let∑∞i=1 xi be a convergent series with de-
creasing terms x1 ≥ x2 ≥ · · · ≥ xn ≥ · · · > 0 and let P be the set of numberswhich can be written in the form
∑i∈J xi for some subset J ∈ Z+. Prove that P is
an interval if and only if
xn ≤∞∑
i=n+1
xi for every n ∈ Z+.
Problem 4.36. (Amer. Math. Monthly) Does there exist a continuous function f :(0,∞) → R such that f (x) = 0 if and only if f (2x) �= 0? What if we require onlythat f be continuous at infinitely many points?
Problem 4.37. (Kvant) Find the smallest number a such that for every real polynomialf (x) of degree two with the property that |f (x)| ≤ 1 for all x ∈ [0, 1], we have|f ′(1)| ≤ a. Find the analogous number b such that |f ′(0)| ≤ b.
Problem 4.38. Let f : R → R be a function for which there exists some constantM > 0 satisfying
|f (x + y)− f (x)− f (y)| ≤ M, for all x, y ∈ R.
Prove that there exists a unique additive function g : R → R such that
|f (x)− g(x)| ≤ M, for all x ∈ R.
Moreover, if f is continuous, then g is linear.
Problem 4.39. (Amer. Math. Monthly) Show that if f is differentiable and if
limt→∞
(f (t)+ f ′(t)
) = 1,
thenlimt→∞ f (t) = 1.
Problem 4.40. (Putnam Competition) Let c be a real number, and let f : R → R bea smooth function of class C3 such that limx→∞ f (x) = c and limx→∞ f ′′′(x) = 0.Show that limn→∞ f ′(x) = limx→∞ f ′′(x) = 0.
34 4 Analysis
Problem 4.41. (Putnam Competition) Prove that the following integral equation hasat most a continuous solution on [0, 1] × [0, 1]:
f (x, y) = 1 +∫ x
0
∫ y
0f (u, v) du dv.
Problem 4.42. (Putnam Competition) Find those λ ∈ R for which the functionalequation ∫ 1
0min(x, y)f (y) dy = λf (x)
has a solution f that is nonzero and continuous on the interval [0, 1]. Find thesesolutions.
Problem 4.43. (Amer. Math. Monthly) Let X be an unbounded subset of the realnumbers R. Prove that the set
AX = {t ∈ R; tX is dense modulo 1}is dense in R.
Problem 4.44. (Putnam Competition) Consider P(z) = zn + a1zn−1 + · · · + an,
where ai ∈ C. If |P(z)| = 1 for all z satisfying |z| = 1, then a1 = · · · = an = 0.
Problem 4.45. Let I ⊂ R be an interval and u, v : I → R smooth functions satisfy-ing the equations
u′′(x)+ A(x)u(x) = 0, v′′(x)+ B(x)v(x) = 0,
where A,B are continuous on I and A(x) ≥ B(x) for all x ∈ I . Assume that v is notidentically zero. If α < β are roots of v, then there exists a root of u that lies withinthe interval (α, β), unless A(x) = B(x), in which case u and v are proportional forα ≤ x ≤ β.
Problem 4.46. (Amer. Math. Monthly) LetV be a finite-dimensional real vector spaceand f : V → R a continuous mapping. For any basis B = {b1, b2, . . . , bn} of V ,consider the set
EB = {z1b1 + · · · + znbn, where zi ∈ Z}.Show that if f is bounded on EB for any choice of the basis B, then f is boundedon V .
Problem 4.47. (Amer. Math. Monthly) It is known that if f, g : C → C are entirefunctions without common zeros then there exist entire functions a, b : C → C suchthat a(z)f (z)+ b(z)g(z) = 1 for all z ∈ C.
1. Prove that we can choose a(z) without any zeros.2. Is it possible to choose both a and b without zeros.
Problem 4.48. (Amer. Math. Monthly) Consider a compact set X ⊂ R. Show that anecessary and sufficient condition for the existence of a monic nonconstant polyno-mial with real coefficients h ∈ R[x] such that |h(x)| < 1 for all x ∈ X is the existenceof monic nonconstant polynomial g(x) ∈ R[x] such that |g(x)| < 2 for all x ∈ X.Prove that 2 is the maximal number with this property.
Part II
Solutions and Comments to the Problems
5
Number Theory Solutions
Problem 1.1. Show that we have Ckn ≡ 0 (mod 2) for all k satisfying 1 ≤ k ≤ n− 1if and only if n = 2β , where β ∈ Z
∗+. Here Ckn denotes the number of combinations,i.e., the number of ways of picking up a subset of k elements from a set of n elements.Known also as the binomial coefficient or choice number and sometimes denoted as(nk
)it is given by the formula
Ckn =(n
k
)= n!
k!(n− k)!
where the factorial n! represents
n! = 1 × 2 × 3 × · · · × n.
Solution 1.1. Denote by expk(m) the maximal exponent of m in k, i.e., the maximalr such that mr divides k. Since Ckn = n!
k!(n−k)! , the exponent of 2 in Ckn has the value
expCkn (2) = expn!(2)− expk!(2)− exp(n−k)!(2),
which is, by hypothesis, strictly positive for all considered k. Now, the exponent of 2in a factorial is given by the following formula:
expm!(2) =∞∑
i=1
[m2i
],
where the brackets denote the integer part. This implies that our claim is equivalentto the inequality
∑
i=1
[ n2i
]−
∑
i=1
[k
2i
]−
∑
i=1
[n− k
2i
]> 0.
Let us now suppose that n = 2β + s, with 1 ≤ s < 2β . If we take k = s, then theexponent above can be calculated as
38 5 Number Theory Solutions
([2β + s
2
]+ · · ·
)−
([ s2
]+ · · ·
)−
([2β
2
]+ · · ·
)= 0,
contradicting our assumptions. Thus, it is necessary that n = 2β .The sufficiency is established as follows. We have the identity (a+b)2 = a2 +b2
in Z/2Z. Using induction on β, one proves that (x + y)2β = x2β + y2β ∈ Z/2Z. In
particular, (x + 1)2α = x2α + 1, and therefore, by identifying the coefficients of the
binomial expansion on the left- and right-hand sides, we obtain the claim.
Comments 1 Using the formula Ckn = Ckn−1 + Ck−1n−1 , we find that Ckn ≡ 1 (mod 2)
for all k ≤ n if and only if n = 2α − 1.
Problem 1.2. Let P = anxn + · · · + a0 be a polynomial with integer coefficients.
Suppose that there exists a number p such that:
1. p does not divide an;2. p divides ai , for all i ≤ n− 1;3. p2 does not divide a0.
Then P is an irreducible polynomial in Z[x].
Solution 1.2. Let ϕ : Z → Z/pZ be the reduction modulo p and let ϕ : Z[x] →Z/pZ[x] be the map consisting of taking the reduction modulo p of all coefficients.This map is actually a homomorphism of rings, meaning that it respects addition andmultiplication of polynomials. We have
ϕ(P (x)) = ϕ(an)xn �= 0.
Assume that P is not irreducible, and so P = P1 · P2. Then ϕ(P ) = ϕ(P1) · ϕ(P2).Moreover, the only way to write ϕ(an)xn ∈ Z/pZ[x] as a product of two polynomialsis βxk · γ xl , where k + l = n and βγ = ϕ(an). In particular, if we write P1(x) =β0 + · · · + βkx
k , P2(x) = γ0 + · · · + γlxl , then ϕ(βk) = β, ϕ(βi) = 0 if i < k,
and ϕ(γi) = γ, ϕ(γi) = 0 if i < l. Eventually, this yields a0 = β0γ0 ≡ 0 (mod p2),which contradicts our assumptions.
Comments 2 This result is known as the Eisenstein criterion. Using this result wecan give an alternative solution to Problem 1.1 above. Assume that Ckn ≡ 0 (mod 2)for k ∈ {1, . . . , n − 1}. Then the Eisenstein criterion shows that the polynomialP(x) = (x + 1)n + 1 is irreducible. This means that Q(x) = P(x − 1) = xn + 1is equally irreducible. On the other hand, if we can write n = 2a(2b + 1), witha, b ∈ Z+, then xn + 1 is divisible by x2a + 1; thus irreducibility of Q(x) impliesthat b = 0. Therefore, n = 2a , as claimed.
Problem 1.3. Given mi, bi ∈ Z+, i ∈ {1, 2, . . . , n} such that gcd(mi,mj ) = 1 forall i �= j , there exist integers x satisfying x ≡ bi (mod mi) for all i. This result isusually known as the Chinese remainder theorem.
5 Number Theory Solutions 39
Solution 1.3. For n = 2, we have m1Z +m2Z = Z. Therefore there exist ai ∈ miZsuch that a1 +a2 = 1. Then x = a1b2 +a2b1 satisfies the conditions of the statement.
Since mi are relatively prime, for each j ∈ {1, 2, . . . , n} there exists yj ∈ Z sat-isfying the conditions yj ≡ 1 (mod mj) and yj ≡ 0 (mod m1 · · ·mj−1mj+1 · · ·mn).Then take x = b1y1 + · · · + bnyn, which satisfies the system of congruences.
Problem 1.4. If gcd(a,m) = 1, then it is a classical result of Euler that we have thefollowing congruence:
aϕ(m)+1 ≡ 0 (mod m),
where ϕ(m) is the Euler totient function, which counts how many positive integerssmaller than m are relatively prime to m. Prove that this equality holds precisely forthose numbers a,m such that for any prime number p that divides a, if pk divides mthen pk also divides a.
Solution 1.4. Let p be a prime number and e the maximal exponent such that pe
divides m. If p divides a, then pe divides a. If p does not divide a, then aϕ(pe) ≡
1 (mod pe), because ϕ(m) is a multiple of ϕ(pe). Therefore aϕ(m) ≡ 1 (mod pe).This implies that pe divides a(aϕ(m) − 1). Hence m divides a(aϕ(m) − 1).
In order to establish the converse, let p be a divisor of a such that pk divides mbut pk does not divide a. Then pk does not divide a(aϕ(m)−1), because obviously, pand aϕ(m) − 1 are relatively prime. Consequently, m is not a divisor of a(aϕ(m) − 1).
Problem 1.5. Let p > 2 be a prime number and let ak ∈ {0, 1, . . . , p2 − 1} denotethe value of kp modulo p2. Prove that
p−1∑
k=1
ak = p3 − p2
2.
Solution 1.5. We have
(p − k)p = (−k)p + C1pp(−k)p−1 + · · · + pp ≡ −kp (mod p2).
Therefore, ap−k = p2 − ak . In particular, summing up these equalities, we obtain∑k−1p−1 ap−k = p2(p − 1)− ∑p−1
k=1 ak , which yields∑p−1k=1 ak = p2(p−1)
2 .
Problem 1.6. If k ∈ Z+, then show that[√k2 + 1 + · · · +
√k2 + 2k
]= 2k2 + 2k,
where [x] denotes the integer part, i.e., the largest integer smaller than x.
Solution 1.6. We have√k2 +m− k = m√
k2+m+k , and if 1 ≤ m ≤ 2k, then m2k+1 ≤
m√k2+m+k ≤ m
2k . Summing up all terms, we obtain
k =2k∑
m=1
m
2k + 1≤
(2k∑
m=1
√k2 +m
)
− 2k2 ≤2k∑
m=1
m
2k= k + 1
2,
whence the claim.
40 5 Number Theory Solutions
Problem 1.7. If n is not a multiple of 5, then P = x4n + x3n + x2n + xn + 1 isdivisible by Q = x4 + x3 + x2 + x + 1.
Solution 1.7. If ξ is a fifth primitive root of unity, then
P(ξ) = ξ4n + ξ3n + ξ2n + ξn + 1 = ξ5n − 1
ξn − 1= 0.
Hence P = (x − ξ)(x − ξ2)(x − ξ3)(x − ξ4)R = Q · R, for some polynomial R.
Comments 3 More generally, x(m−1)n + · · · + 1 is divisible by xm−1 + · · · + 1 ifgcd(m, n) = 1.
Problem 1.8. Let p be an odd prime and k ∈ Z+. Show that there exists a perfectsquare the last k digits of whose expansion in base p are 1.
Solution 1.8. Use induction on k. It is obvious for k = 1. Let us assume that x2 =1 + p + · · · + pk−1 + (apk + · · · ) and gcd(x, p) = 1. Then we have
(x + cpk)2 = x2 + 2cxpk + c2p2k = 1 + p + · · · + pk−1 + (a + 2cx)pk + · · · .If p is odd, then gcd(2x, p) = 1. Moreover, the congruence 2xc + a ≡ 1 (mod p)has at least one solution c ∈ {0, 1, . . . , p − 1}. If c0 is a solution, then (x + c0p
k)2
has its last k + 1 digits equal to 1.
Comments 4 The result can be extended to higher exponents r ≥ 2 and primenumbers p that do not divide r , or more generally to p such that gcd(p, r) = 1.Moreover, the last digits can be arbitrarily prescribed.
Problem 1.9. Any natural number greater than 6 can be written as a sum of twonumbers that are relatively prime.
Solution 1.9. For odd n ≥ 7, we can write n = 2k + 1 = k + (k + 1), wheregcd(k, k+1) = 1. Consider now n = 2m, wherem ≥ 3. Recall Chebyshev’s theoremthat for anym ≥ 2 there exists a prime number p such thatm < p < 2m. Therefore,we have 0 < 2m − p < m. Further, gcd(p, 2m − p) = 1, since 2m > p > m andgcd(m, p) = 1.
Problem 1.10. Prove that it is impossible to extract an infinite arithmetic progressionfrom the sequence S = {1, 2k, 3k, . . . , nk, . . .}, where k ≥ 2.
Solution 1.10. Let r be the ratio of an arithmetic progression A ⊂ S. If n > r , then(n + 1)k − nk ≥ 2n + 1 > r . Thus the difference between consecutive terms of Sgrows higher than r , and hence A is finite.
Comments 5 The result is also true when we replace S with the sequence an+2 =pan+1 + qan, for 1 ≤ p ≤ q + 1.
Problem 1.11. Prove that ba−j+1 divides Cjba if a, b ≥ 2, j ≤ a + 1.
5 Number Theory Solutions 41
Solution 1.11. An easy induction on n shows that bn ≥ n+ 1, and hence ba ≥ j . Wewrite the combinations (also called binomial coefficients) as
Cjba = 1
j !
j−1∏
i=0
(ba − i).
Let i = brm, where gcd(m, b) = 1, so that ba − i = br(ba−r −m). This shows thati and ba − i are divisible by the same power of b. Therefore, the exponent of b in Cjbaequals the result of subtracting from a the exponent of b in j !. The latter is at mostj − 1 because bj−1 ≥ j , and hence the claim.
Problem 1.12. Solve in integers the following equations: (1) x2 = y2 + y3; (2)x2 + y2 = z2.
Solution 1.12. 1. The plane curve determined by the first equation has the followingform:
x
y
0
DT
P
We will look for a parameterization of this curve by projecting from the origin 0onto the line D given by y = 1. Thus, to each point P of the curve we associate thepoint T , which is the intersection of the half-line 0P with the lineD. Denote by t thenatural parameter of D. This amounts to setting y = 1
tx within the equation, which
immediately yields x = t3 − t and y = t2 − 1.If x, y ∈ Z, then t ∈ Q. Moreover, t2 = y + 1 ∈ Z, but a rational number whose
square is an integer must be an integer itself. Thus the integer solutions are given bythe family x = t3 − t and y = t2 − 1, where t ∈ Z.
2. Dividing by z2, we reduced the problem of finding the rational solutions of theequation x2 + y2 = 1. This is a circle of unit radius in the plane. We now use thestereographic projection from the point P(−1, 0) onto the vertical line x = 1 in orderto find a more-convenient parameterization of the circle.
P Q
XT
42 5 Number Theory Solutions
If t is the new parameter on the vertical line, then x2 + y2 = 1 and x+12 = y
t. This
yields x = 1−s2
1+s2 and y = 2s1+s2 , where s = t
2 .
Therefore the integer solutions of our equation are given by x = (r2 − s2)v,y = 2vrs, and z = (r2 + s2)v, where s, r, v ∈ Z. The primitive solutions, for whichgcd(x, y, z) = 1, are those for which v = 1 and gcd(r, s) = 1. The triples (x, y, z)are called Pythagorean triples.
Comments 6 The two Diophantine equations above correspond to rational algebraiccurves, i.e., to curves in the plane that admit a natural parameterization by means ofrational functions. This situation is quite exceptional.
Comments 7 The Pythagorean equation x2 +y2 = z2 can be generalized by addingmore variables and still keeping its rational behavior to
x21 + x2
2 + · · · + x2n+1 = z2, n ≥ 1.
The general (primitive) solution of this equation in integers can be obtained by thesame method as above and reads
x1 = s21 + s2
2 + · · · + s2n − u2,
x2 = 2s1u,
x3 = 2s2u,
......
xn+1 = 2snu,
z = s21 + s2
2 + · · · + s2n + u2,
for arbitrary integer parameters s1, s2, . . . , sn, u ∈ Z.
Comments 8 Another famous generalization of the Pythagorean equation is
xk + yk = zk, k ≥ 3,
which was conjectured by Fermat to have no nontrivial solutions for k ≥ 3. Severalparticular cases were solved over the years (k = 4 by Fermat, k = 3 by Euler,k = 5, 14 by Dirichlet, k = 7 by Lamé, etc). However, it took more than threehundred years for the conjecture be eventually settled by A.Wiles in 1995.
Comments 9 Euler considered the more general equation
xk1 + xk2 + · · · + xkn = zk.
He conjectured that there are no nontrivial solutions unless n ≥ k and proved it forn = 3. However, Lander and Parkin found counterexamples for n = 5, for example,
275 + 845 + 1105 + 1335 = 1445,
5 Number Theory Solutions 43
and later, N. Elkies settled the case n = 4 by finding another counterexample:
2 682 4404 + 15 365 6394 + 18 796 7604 = 20 615 6734.
A subsequent computer search by R. Frye found the following minimal solution forn = 4 (which is unique in the range z ≤ 1 000 000):
95 8004 + 217 5194 + 414 5604 = 422 4814.
See also
• N.D. Elkies: On A4 + B4 + C4 = D4, Math. Comp. 51 (1988), 184, 825–835.• J.L. Lander and T.R. Parkin: A counterexample to Euler’s sum of powers conjec-
ture. Math. Comp. 21 (1967), 101–103.
Problem 1.13. Let a, b, c, d ∈ Z+ be such that at least one of a and c is not a perfectsquare and gcd(a, c) = 1. Show that there exist infinitely many natural numbers nsuch that an + b, cn + d are simultaneously perfect squares if one of the followingconditions is satisfied:
1. b and d are perfect squares;2. a + b, c + d are perfect squares;3. a(d − 1) = c(b − 1).
Moreover, there do not exist such numbers if a = 1, b = 0, c = 4k2 − 1, d = 1.
Solution 1.13. There exists n such that an + b, cn + d are perfect squares iff theequation
ax2 − cy2 = ad − bc
has integer solutions. In fact,√cn+ d,
√an+ b are solutions of the equation. Con-
versely, if (x, y) is a solution, then
a(x2 − d) = c(y2 − b).
Since gcd(a, c) = 1, we find that c divides x2−d, while a divides y2−b. In particular,we obtain that x2 − d = kc and y2 − b = ka.
Further, ac is not a perfect square and so the Pell equation
u2 − acv2 = 1
has infinitely many solutions (un, vn). If (x, y) is a solution of our equation, then
xn = unx + dvny, yn = uny + vnx
form an infinite sequence of solutions. Thus it suffices to determine one solution forthe equation in order to find infinitely many.
The first two cases are immediate, and the third corresponds to the solution (1, 1).Finally, note that the Pell-type equation x2 − (4k2 − 1)y2 = −1 has no integer
solutions, by modulo 4 considerations.See also:
44 5 Number Theory Solutions
• T. Andreescu and D. Andrica, Quadratic Diophantine Equations, Springer Mono-graphs in Math., 2006.
Problem 1.14. If N = 2 + 2√
28n2 + 1 ∈ Z for a natural number n, then N is aperfect square.
Solution 1.14. If√
28n2 + 1 = k, then N = 2 + 2k. If N ∈ Z, then k ∈ 12 Z. But k
is the square root of an integer, and whenever such a square root is a rational number,then it is actually an integer. This proves that k ∈ Z. Now, 28n2 = (k − 1)(k + 1),and therefore k is odd, k = 2t + 1. Thus, 7n2 = t (t + 1). We have two cases:
1. If 7 divides t + 1, then, t = 7s − 1 and so (7s − 1)s = n2, for s ∈ Z. Butgcd(s, 7s − 1) = 1 and thus 7s − 1 = n2
1 and s = n22, where n1n2 = n. Moreover,
any square, in particular n21, is congruent to 1, 2, or 4 (mod 7). In particular, it cannot
be equal to 7s − 1.2. If 7 divides t , i.e., t = 7s, then n2 = s(7s + 1). Since gcd(s, 7s + 1) = 1, we
have 7s + 1 = n21 and s = n2
2. This implies that
N = 2 + 2k = 2 + 2(14n22 + 1) = 4(7n2
2 + 1) = 4(7s + 1) = 4n21.
Problem 1.15. Let n ≥ 5, 2 ≤ b ≤ n. Prove that[(n− 1)!
b
]≡ 0 (mod b − 1).
Solution 1.15. We have four cases to consider:1. If b < n, then (n−1)! is divisible by b(b−1) and therefore (n−1)!
bis an integer
divisible by b − 1.2. Suppose b = n, where n is not prime and not the square of a prime number,
hence n = rs, where 1 < r < s < n. Since gcd(n, n − 1) = 1 and s < n − 1, wederive that (n − 1)! contains the factors r, s, and n − 1, and therefore it is divisibleby rs(n− 1) = b(b − 1).
3. If b = n is the square of a prime, i.e., n = p2, then p ≥ 3 and hence1 < p < 2p < 2p2 −1 = n−1. This implies that (n−1)! is divisible by the productof p, 2p, and n− 1, i.e., by 2p2(n− 1) = 2b(b − 1).
4. If b = n is a prime number p, then by Wilson’s theorem we have (p−1)!+1 ≡0 (mod p),[(p − 1)!
p
]=
[(p − 1)! + 1
p− 1
p
]= (p − 1)! + 1
p−1 = (p−1)
((p − 2)! − 1
p
).
Now gcd(p, p − 1) = 1, so p − 1 divides[(p−1)!p
].
Problem 1.16. Prove that for every natural number n, there exists a natural numberk such that k appears in exactly n nontrivial Pythagorean triples.
Solution 1.16. Let us show that 2n+1 appears in exactly n triples. If n = 0, it amountsto saying that 2 does not appear in any such triple, which is immediate.
5 Number Theory Solutions 45
Let us now use induction on n. All primitive Pythagorean triples are given by
x = u2 − v2, y = 2uv, z = u2 + v2,
where gcd(u, v) = 1 and u, v are not both odd and u > v.If u, v are both odd, then y = 2uv = 2n+1 and therefore u = 2n and v = 1.
The nonprimitive triples in which 2n+1 appears are those divisible by 2, and thus theyare in bijection with the Pythagorean triples in which 2n appears. By the recurrencehypothesis we have n such triples. Therefore 2n+1 appears in n nonprimitive triplesand one primitive triple, and thus in n+ 1 such triples.
Problem 1.17. Let n, q ∈ Z+ be such that all prime divisors of q are greater than n.Show that
(q − 1)(q2 − 1) · · · (qn−1 − 1) ≡ 0 (mod n!).
Solution 1.17. Let p ≤ n, prime. Since p is not a divisor of qk , we have qk(p−1) ≡1 (mod p), for all k. Therefore at least
[(n−1)(p−1)
]factors from the left-hand side are
divisible by p.Let us now compute the exponent of p in n!, which is
expn!(p) =∞∑
k=1
[n
pk
]<
∞∑
k
n
pk= n
p − 1.
Now expn!(p) ∈ Z, and so the strict inequality above implies that expn!(p) ≤[n−1p−1
].
Thus the left-hand side is divisible by pexpn!(p). Since this holds for all p ≤ n, theclaim follows.
Comments 10 Let q = pm, where p > n is a prime number. LetG denote the groupof nonsingular matrices over the Galois field with q elements. Then the order of thegroupG is |G| = qn(n−1)/2(q−1) · · · (qn−1). Let nowG∗ be the subgroup generatedby the diagonal matrices and their permutations. One finds that |G∗| = (q − 1)nn!.Since the order of a subgroup divides the order of the larger group, we derive
(q − 1) · · · (qn − 1) ≡ 0 (mod (qn − 1)nn!),
improving the relation above.
Problem 1.18. Every natural number n ≥ 6 can be written as a sum of distinct primes.
Solution 1.18. Let n ≥ 15. One knows by Chebyshev’s theorem that there exists aprime p between n
2 and n. Then 0 < n− p < n2 . Further, there exists a prime q such
that n−p2 < q < n−p, and the rest n−p− q satisfies therefore 0 < n−p− q < n4 .
We continue this process until there remains either a prime number or a number fromthe set {7, 8, . . . , 14}. Since 7 = 5 + 2, 8 = 5 + 3, 9 = 7 + 2, 10 = 7 + 3, 11 = 11,12 = 7 + 5, 13 = 13, and 14 = 2 + 5 + 7, the claim follows.
46 5 Number Theory Solutions
Problem 1.19. Every number n ≥ 6 can be written as a sum of three numbers whichare pairwise relatively prime.
Solution 1.19. By Chebyshev’s theorem , there exists a primep such that n2 < p < n.Note next that a natural number m such that m is divisible by all prime numbers
strictly smaller than m must be 2. In fact, otherwise, again by Chebyshev’s theorem,there exists a prime bigger than m
2 , which cannot divide m. Thus there exists a primeq such that n− p is not divisible by q, and so gcd(n− p, q) = 1.
If gcd(p, n−p−q) > 1, then n−p−q = ap, which cannot happen, sincep > n2
and a �= 0, since n − p − q > 0. If gcd(q, n − p − q) > 1, then n − p − q = aq,which contradicts our choice of q. Thus p, q, n− p − q satisfy the claim.
Problem 1.20. Find all pairs of integers (m, n) such that
Cnm = 1984,
where Cnm = m!n!(m−n)! denotes the usual binomial coefficient.
Solution 1.20. We consider 0 ≤ n ≤ m2 . We have 1984 = 26 · 31. Since 1984 =
(n+1)(n+2)···(m−1)m(m−n)! , then either 31 divides (n+ 1)(n+ 2) · · · (m− 1)m (and does not
divide (m − n)!), so that m ≥ 31, or else 31 also divides (m − n)! and hence againm ≥ 31.
If n ≥ 3, then Cnm ≥ C3m ≥ C3
31 = 16 (29 · 30 · 31) > 64 · 31 = 1984. Thus
n ∈ {0, 1, 2}.Finally, if n = 1, then m = 1984, while C2
m = 1984 does not have any solutionsin natural numbers.
Thus, the solutions are (1984, 1), (1984, 1983).
Comments 11 More generally, if N has a prime divisor p > k(k√N + 1
), k < p
2 ,
then the equation Cnm = N has at most 2k solutions. One derives, furthermore, thatthe number F(N) of such solutions can be estimated from above:
F(N) = O(logN/ log logN).
Recall that F(N) = O(g(N)) if limN→∞ F(N)g(N)
< ∞. It is conjectured that F(N) isuniformly bounded, independently of N .
Problem 1.21. Find the setA consisting of natural numbers n that are divisible by allodd natural numbers a with a2 < n.
Solution 1.21. Let p1 = 3, p2, . . . , pm be the firstm odd primes. Choose k such that(2k − 1)2 < n ≤ (2k + 1)2 and m maximal such that pm ≤ 2k − 1. By hypothesis,pm+1 > 2k + 1 and hence, n < p2
m+1.Assume that n is in A. Then the pi divide n for all i ≤ m, and thus p1p2 · · ·pm
divides n. Moreover, the Bonse–Pósa inequality states that
p1 · · ·pm > p2m+1, m ≥ 4.
5 Number Theory Solutions 47
Thus m ≤ 3.If m = 3, then n ≤ 169 and 105 divides n. However, n = 105 does not belong to
A, since it is not divisible by 9.If m = 2, then n ≤ 49 and n is divisible by 15.If m = 1, then n ≤ 25 is divisible by 3.If m = 0, then n ≤ 9.The answer is thereforeA = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 15, 18, 21, 24, 30, 45}.
Comments 12 Panaitopol recently gave an elementary proof of a more-generalBonse–Pósa inequality p1p2 · · ·pn > p
n−π(n)n+1 , for n ≥ 2, which implies that
p1p2 · · ·pn > pkn+1 for n > 2k. This has been further improved by Alzer and Berg.
• L. Panaitopol: An inequality involving prime numbers, Univ. Beograd. Publ. Elek-trotehn. Fak. Ser. Mat. 11 (2000), 33–35.
• H. Alzer, C. Berg: Some classes of completely monotonic functions II, The Ra-manujan Journal 11(2006), 225–248.
Problem 1.22. Prove that 21092 − 1 is divisible by 10932.
Solution 1.22. Set p = 1093 and thus p2 = 1194649. Observe that37 = 2187 = 2p + 1,314 ≡ 4p + 1 (mod p2),214 = 16348 = 15p − 11, 228 ≡ −330p + 121 (mod p2),32 · 228 ≡ −2970p + 1089 ≡ −1876p − 4 (mod p2), and therefore32 · 226 ≡ −469p − 1 (mod p2). Therefore we infer that314 · 2182 ≡ −(489p + 1)7 ≡ −3283p − 1 ≡ −4p − 1 ≡ −314 (mod p2),
and using the fact gcd(314, p2) = 1, we obtain 2182 ≡ −1 (mod p2), yielding21092 − 1 ≡ 0 (mod 10932).
Problem 1.23. Let n ≥ 0, r > 1, and 0 < a ≤ r be three integers. Prove that thenumber n, when written in base r , has precisely
∞∑
k=1
[nr−k + ar−1
]−
[nr−k
]
digits that are greater than or equal to r − a.
Solution 1.23. We write n = n1 + n2r + n3r2 + · · · in base r . Then the digits nj are
computed by the formulas
n1 = n−[nr
]r, n2 =
[nr
]−
[ nr2
]r, . . . , nk =
[ n
rk−1
]−
[ nrk
]r.
For a fixed k, we have nk ≥ r − a if and only if[
nrk−1
]−
[nrk
]r ≥ r − a, which is
equivalent to asking that[
1r
([nrk−1
]+ a
)]≥
[nrk
]+ 1.
Now let n = mrk−1 + s, where 0 ≤ s < rk−1. Then
48 5 Number Theory Solutions[
1
r
([ n
rk−1
]+ a
)]=
[mr
+ a
r
]=
[mr
+ s
rk+ a
r
]=
[ nrk
+ a
r
]
and therefore nk ≥ r − a if and only if[n1rk
+ ar
]>
[nrk
]. However, we have
−[ nrk
]+
[ nrk
+ a
r
]=
{0, if nk < r − a,
1, if nk ≥ r − a.
Thus, the total number of digits greater than r − a is
∞∑
k=1
([ nrk
+ a
r
]−
[ nrk
])
as claimed.
Comments 13 The same argument shows that given a sequence ak ∈ {1, 2, . . . , r},the number of digits nk of n written in base r that satisfy the inequalities
nk ≥ r − ak
is ∞∑
k=1
([ nrk
+ ak
r
]−
[ nrk
]).
Problem 1.24. Find a pair (a, b) of natural numbers satisfying the following proper-ties:
1. ab(a + b) is not divisible by 7;2. (a + b)7 − a7 − b7 is divisible by 77.
Solution 1.24. Since a, b, a + b �≡ 0 (mod 7), there exists the inverse a−1 of a inZ/77
Z; moreover, k = ba−1 �≡ 0,−1 (mod 77). Then 77 divides (a−1)7((a + b)7 −a7 − b7) = (1 + k)7 − 1 − k7. Developing the latter, we find that
7k(k5 + 3k4 + 5k3 + 5k2 + 3k + 1) ≡ 0 (mod 77);thus
(k + 1)(k4 + 2k3 + 3k2 + 2k + 1) ≡ 0 (mod 76).
Since k �≡ −1 (mod 77), we obtain
(k2 + k + 1)2 ≡ 0 (mod 76),
and so we have to solve in integers the equation k2 +k+1 = 73s. The discriminant isD = 4·73s−3, which has to be a squarep2. We need then to findp such that 73 dividesp2 + 3. We note that 72 divides p2 + 3 only if either p = 12 + 49q or p = 37 + 49q.By replacing above, we find in the first case p2 + 3 = 49(24q + 3) ≡ 0 (mod 343),which yields q ≡ −1 (mod 7) and thus k = −19 + 343r . The second case yields thesolutions k = 18 + 343r .
In particular, (1, 18) and (1, 324) are solutions for the problem.
5 Number Theory Solutions 49
Problem 1.25. Let 0 < a < b < c < d be odd integers such that
1. ad = bc
2. a + d = 2k, b + c = 2m, for some integers k and m.
Prove that a = 1.
Solution 1.25. Let ab
= cd
= xy
, where gcd(x, y) = 1. One obtains then, from theother equations, the value
a = x2m(y − 2k−mx)y2 − x2 .
Since a is odd, we have y2 − x2 = 2ms for some integer s. Recall that x, y areodd, since they divide a and b, respectively. We obtain, then, the following types ofsolutions:
1.
{y = 2m−2s1 + s2,
x = 2m−2s1 − s2,
2.
{y = s1 + 2m−2s2,
x = s1 − 2m−2s2,
where the si are both integers. Moreover, when replacing x and y by these valuesin the expression of a, we obtain that s1 must divide 2k−m + 1 and s2 must divide2k−m − 1. Further, we have the equivalences
k > m iff a + d > b + c iff a + bc
a> b + c iff (a − b)(a − c) > 0.
The inequality a + bca
≤ 1 + bc implies that k < 2m− 2.Now, in the first type of solutions, we have s1 > 2m−2s2, and s1 divides 2k−m+1,
hence s2 = 1, s1 = 2k−m + 1, k = 2m− 2.In the second case, we have y ≤ b < 2m−1 and thus s1 < 2, which yields s1 = 1.
Since 3y > 2m−1, we find that x = a, y = b, and replacing in the expression of a,we obtain s2 = 2m−2 − 1, so that a = 1.
Problem 1.26. Find those subsets S ⊂ Z+ such that all but finitely many sums ofelements from S (possibly with repetitions) are composite numbers.
Solution 1.26. Let S∗ = {x = ∑ai∈S ai} be the semigroup generated by S. We prove
that S∗ contains only composite numbers iff there exists a prime number p, withp /∈ S, such that p divides all elements of S. This condition is obviously sufficient.In order to prove the converse, let us assume the contrary, i.e., that gcd(S) = 1. Thenthere exists a finite subset T = {c1, . . . , cn} ⊂ S such that gcd(T ) = 1. This impliesthat there exist integers xi ∈ Z such that
n∑
i=1
cixi = 1.
50 5 Number Theory Solutions
Separating the terms with xi > 0 and those with xi < 0, we obtain two elementsa, b ∈ S∗ with the property that a − b = 1. In particular, gcd(a, b) = 1. FromDirichlet’s theorem, the arithmetic progression {a+kb}k∈Z+ contains infinitely manyprimes; but a + kb ∈ S∗, contradicting our assumptions. The claim follows.
Comments 14 Another proof can be given by using the Chinese remainder theoremand showing that if S is such that gcd(S) = 1, then S∗ is Z+ −A, where A is a finitesubset. The problem of determining the maximal element of A is called the Frobeniuscoin problem. If S = {a1, a2}, then the maximal number is F(S) = a1a2 − a1 − a2,as was proved by Sylvester. If card(S) = 3, the problem was solved by Selmer andBeyer, but for card(S) ≥ 4, there is no closed formula for F(S). Erdos and Grahamproved that
F({a1 < a2 < · · · < an}) ≤ 2an−1
[ann
]− an,
and Brauer that
F({a1 < a2 < · · · < an}) ≤n∑
i=1
ai
(gcd(a1, a2, . . . , ai)
gcd(a1, a2, . . . , ai+1)− 1
).
Curtis proved that there is no closed formula expressing F(S) for n ≥ 4 that involvesonly polynomials.
• A. Brauer: On a problem of partitions, Amer. J. Math. 64 (1942), 299–312.• F. Curtis: On formulas for the Frobenius number of a numerical semi-group, Math.
Scand. 67 (1990), 2, 190–192.• P. Erdos and R.L. Graham: On a linear Diophantine problem of Frobenius, Acta
Arith. 21 (1972), 399–408.• E.S. Selmer and Ö. Beyer: On the linear Diophantine problem of Frobenius in
three variables, J. Reine Angew. Math. 301 (1978), 161–170.
Problem 1.27. Prove that for any natural number n > 1, the number 2n − 1 does notdivide 3n − 1.
Solution 1.27. SetAn = 2n−1 and Bn = 3n−1, where n > 1. For even n, 3 dividesAn, while 3 does not divide Bn, and so An does not divide Bn.
Now let n > 1 be odd, of the form n = 2m−1. Using 24 ≡ 22 (mod 12), we inferAn ≡ −5 (mod 12). Since any prime greater than three is congruent to either ±5, or±1 modulo 12, there is at least one prime divisor p ofAn such that p ≡ ±5 (mod 12).IfAn dividesBn, thenp divides 3Bn and thus 3 ≡ 3n+1 ≡ 32m (modp), which impliesthat 3 is a quadratic residue modulo p. But this contradicts the quadratic reciprocitytheorem (an easy consequence of which is the fact that 3 cannot be a quadratic residuemodulo p), since p ≡ ±5 (mod 12).
Problem 1.28. Let un be the least common multiple of the first n terms of a strictlyincreasing sequence of positive integers. Prove that
∞∑
n=1
1
un≤ 2.
5 Number Theory Solutions 51
Find a sequence for which equality holds above.
Solution 1.28. Let ak be the increasing sequence, and uk = lcm(a1, a2, . . . , ak). Wehave ∞∑
k=1
1
uk=
∞∑
k=1
1
lcm(a1, a2, . . . , ak)≤ 1
a1+
∞∑
k=2
1
lcm(ak, ak+1)
= 1
a1+
∞∑
k=2
gcd(ak, ak+1)
akak+1≤ 1
a1+
∞∑
k=2
ak+1 − ak
akak+1
= 1
a1+
∞∑
k=2
(1
ak− 1
ak+1
)= 2
a1≤ 2.
Examples of sequences for which equality holds are ak = 2k−1, and a1 = 1, a2 =2, a3 = 6, ak = 3 · 2k−2, for k ≥ 4.
Problem 1.29. Let ϕn(m) = ϕ(ϕn−1(m)), where ϕ1(m) = ϕ(m) is the Euler totientfunction, and set ω(m) the smallest number n such that ϕn(m) = 1. If m < 2α , thenprove that ω(m) ≤ α.
Solution 1.29. If we consider the minimal j with ϕj (m) = 1, then ϕj−1(m) = 2.Also, we have ϕ1(m) < m < 2α , but ϕ1(m) is an even number and thus ϕ2(m) ≤12ϕ1(m) < 2α−1. By induction, we obtain ϕj−1(m) < 2α−(j−2), whence the claim.For even m, we have ω(m) ≤ α − 1.
Problem 1.30. Let f be a polynomial with integer coefficients andN(f ) = card{k ∈Z; f (k) = ±1}. Prove thatN(f ) ≤ 2 + deg f , where deg f denotes the degree of f .
Solution 1.30. Set N+(f ) = card{k; k ∈ Z, f (k) = 1} and, respectively, N−(f ) =card{k; k ∈ Z, f (k) = −1}. We claim first that either N+(f ) ≤ 2 or N−(f ) ≤ 2.Assume the contrary. Then there exist a, b, c, d, e, g distinct integers such that
f (a) = f (b) = f (c) = −1 and f (d) = f (e) = f (g) = +1.
Thus we can write
f (x) = Q(x)(x − a)(x − b)(x − c)− 1 = R(x)(x − d)(x − e)(x − g)+ 1,
whereQ and R are polynomials with integer coefficients. Consequently, we have thepolynomial identity
(x − a)(x − b)(x − c)Q(x) = (x − d)(x − e)(x − g)R(x)+ 2.
Let x take the values a, b, and c, respectively. Then we find that a − d, a − e, a − g
divide 2 and therefore a − d, a − e, a − g ∈ {1,−1, 2,−2}. In a similar way, b −d, b − e, b − g ∈ {1,−1, 2,−2} and c − d, c − e, c − g ∈ {1,−1, 2,−2}. We canassume, from symmetry, that a < b < c and g < e < d. We have then the strictinequalities a − d < b− d < e− d < c− e < c− g, and hence at least five distinctdifference numbers that belong to {1,−1, 2,−2}, which is absurd. Thus our startingassumption is false.
In particular, if N+(f ) ≤ 2, then N(f ) ≤ N(f )+ 2 ≤ deg f + 2.
52 5 Number Theory Solutions
Comments 15 The result can be improved as follows:
N(f )− deg f =
⎧⎪⎪⎨
⎪⎪⎩
1, if deg f = 1,≤ 2, if deg f ∈ {2, 3},≤ 1, if deg f = 4,≤ 0, if deg f ≥ 5.
The solution is analogous. The equality for degree 2 can be attained, for instance, bytaking f (x) = x2 − (2k + 3)x + k2 + 3k + 1.
Problem 1.31. Prove that every integer can be written as a sum of 5 perfect cubes.
Solution 1.31. We have
6n = 03 + (n+ 1)3 + (n− 1)3 + (−n)3 + (−n)3,and this is written as a sum of four perfect cubes. Thus
1. 6n+ 1 = 6n+ 13,2. 6n+ 2 = 6(n− 1)+ 23,3. 6n+ 3 = 6(n− 4)+ 33,4. 6n+ 4 = 6(n+ 2)+ (−2)3,5. 6n+ 5 = 6(n+ 1)+ (−1)3.
Comments 16 Observe that we did not require the cubes to be positive. It is stillunknown whether every integer is the sum of four cubes, although three cubes do notsuffice. For instance, numbers that are congruent to ±4 modulo 9 cannot be writtenas sums of three cubes, and moreover, it is not yet known whether they can be writtenas sums of four cubes. If we are looking for positive cubes, then the minimal numberneeded is 9, this being a particular case of the Waring problem. Notice that anysufficiently large number is a sum of seven positive cubes. The proof is more involved;for instance, see:
• G.L. Watson: A proof of the seven cube theorem. J. London Math. Soc. 26 (1951),153–156.
Problem 1.32. If n ∈ Z, then the binomial coefficient Cn2n = (2n)!(n!)2
has an evennumber of divisors.
Solution 1.32. The number of divisors of x, usually denoted by τ(x), is odd if andonly if x is a perfect square. In fact, once we have a divisor d , there is a complementarydivisor associated with it, namely x
d. These divisors are distinct unless x is a square
and the divisor is its square root.Further, it is known that for every natural number n, there exists a prime p satis-
fying n < p ≤ 2n, and therefore 2n < 2p ≤ 4n. Writing Cn2n = (2n)!n!n! , we find that
the prime p appears with exponent 1 in the development of Cn2n and thus it cannot bea perfect square.
5 Number Theory Solutions 53
Problem 1.33. Prove that every n ∈ Z+ can be written in precisely k(n) differentways as a sum of consecutive integers, where k(n) is the number of odd divisors of ngreater than 1.
Solution 1.33. We have x + (x + 1)+ (x + 2)+ · · · + (x + y) = 12 (y + 1)(2x + y).
Therefore, we must find the number of natural solutions (x, y) of the Diophantineequation 1
2 (y + 1)(2x + y) = n. We have two cases to consider:1. If y = 2z is even, then (2z + 1)(x + z) = n. If we let u = 2z + 1, then u
is an odd divisor of n. There is also associated the complementary divisor v, so thatuv = n. Now v = x+ z, and thus we find that z = 1
2 (u− 1) and x = 12 (2v− u+ 1).
Therefore (x, z) is an acceptable solution if and only if 2v − u ≥ 1.2. If y = 2z − 1 is odd, then z(2x + 2z − 1) = n. Let u = 2x + 2z − 1, which
is an odd divisor of S, and let v be the complementary divisor: uv = n. Then z = v;hence x = 1
2 (u − 2v + 1). This implies that (x, z) is an acceptable solution if andonly if 2v − u ≤ −1.
Now, given u > 1, an odd divisor of n, then either the first inequality or the second(but not both) could hold. Therefore, the number of solutions is that claimed.
Problem 1.34. Let π2(x) denote the number of twin primes p with p ≤ x. Recallthat p is a twin prime if both p and p + 2 are prime. Show that
π2(x) = 2 +∑
7≤n≤xsin
(π
2(n+ 2)
[n!
n+ 2
])sin
(π
2n
[(n− 2)!
n
])
for x > 7.
Solution 1.34. If n > 5 is composite, then (n−2)!n
is an even integer, and therefore
sin(π2 n
[(n−2)!n
])= 0.
If p is prime, then by Wilson’s theorem, (p − 2)! ≡ −(p − 1)! ≡ 1 (mod p),which implies that [
(p − 2)!
p
]= (p − 2)! − 1
p.
If p > 5, then 4 divides (p − 2)!, and therefore
sin
(π
2p
[(p − 2)!
p
])= sin
(π2
[(p − 2)! − 1])
= −1.
Consequently, the nth term of the sum is zero if at least one among n, n + 2 iscomposite, and 1 if both n and n+ 2 are prime. Finally, one adds 2 units in order tocount the pairs (3, 5), (5, 7).
Problem 1.35. 1. Find all solutions of the equation 3x+1 + 100 = 7x−1.2. Find two solutions of the equation 3x + 3x
2 = 2x + 4x2, and prove that there are
no others.
54 5 Number Theory Solutions
Solution 1.35. 1. x = 4 is a solution. If x < 4, then 3x+1 + 100 > 7x−1, and ifx > 4, then 3x+1 + 100 < 7x−1. Therefore, this solution is unique.
2. Let ρ(x) = 4x2 −3x
2
3x−2x . We have
ρ′(x) = (2 · 4x2
ln 4 − 2 · 3x2
ln 3)x(3x − 2x)− (4x2 − 3x
2)(3x ln 3 − 2x log 2)
(3x − 2x)2.
Therefore ρ′(x) ≥ (4x2 −3x
2)(3x−2x)(x−1) ln 3 = μ(x). Next, we haveμ(x) = 0
if and only if x ∈ {0, 1}. These values are isolated roots of ρ′, and thus ρ is strictlyincreasing forx ∈ R+. Therefore, the only two solutions of the equation arex ∈ {0, 1},because, for other values of x, we have ρ(x) > 1.
Problem 1.36. Let σ(n) denote the sum of the divisors of n. Prove that there existinfinitely many integersn such thatσ(n) > 2n, or, even stronger, such thatσ(n) > 3n.Prove also that σ(n) < n(1 + log n).
Solution 1.36. 1. Recall that if n = pa11 . . . p
akk is the factorization of n into prime
factors, then we have the formula σ(n) = ∏ki=1
pai+1i −1pi−1 . Observe next that n = 2p3q
satisfies the inequality σ(n) > 2n for p, q > 2. Moreover, if n = 2p3q5s , thenσ(n) > 3n.
2. We have σ(n) = ∑d/n d; hence
σ(n)
n=
∑
d/n
d
n=
∑
d/n
1
d=
∑
d/n
1
d< 1 + 1
2+ 1
3+ · · · + 1
n< 1 + log n.
Therefore σ(n) < n(1 + log n).
Problem 1.37. Let ai be natural numbers such that gcd(ai, aj ) = 1, and ai are notprime numbers. Show that:
1
a1+ · · · + 1
an< 2
Solution 1.37. Each ai has at least one proper prime divisor pi . Thus one can writeai = piqisi , wherepi and qi are, not necessarily distinct, primes. Now, gcd(ai, aj ) =1 implies that pi, qi �= pj , qj . Therefore, the n numbers min(pi, qi) are distinct. Wederive that
∑ 1
ai≤
∑ 1
piqi≤
∑ 1
(min(pi, qj ))2≤
∑
k≥1
1
k2 = ζ(2) = π2
6< 2.
Comments 17 We denoted above by ζ the Riemann zeta function, defined by
ζ(s) =∑
k≥1
k−s .
One knows the values of the zeta function at specific points. Let Bn denote the nthBernoulli number, determined by the Taylor expansion
5 Number Theory Solutions 55
x
ex − 1=
∞∑
n=0
Bn
n!xn.
The Bernoulli numbers are rational for all n and have the property that B2n+1 = 0 ifn ≥ 1. Their first values are B0 = 1, B1 = − 1
2 , B2 = 16 , B4 = − 1
30 , B6 = 142 . Using
contour integration, one finds the following expression for the Riemann zeta functionin terms of Bernoulli numbers:
ζ(2n) = (−1)n−1 (2π)2n
2(2n)!B2n
for all natural n. In particular, ζ(n) is transcendental for even n. It is considerablymore difficult to show the irrationality at odd values. The arithmetic nature of the zetavalues at odd positive integers has remained a mystery since Euler’s time even thoughone now conjectures that the numbersπ, ζ(3), ζ(5), . . . are algebraically independentover the rationals. Apéry finally proved in 1979 that ζ(3) is irrational. Recent progresswas made by Rivoal in 2000, who showed that there are infinitely many integers nsuch that ζ(2n+ 1) is irrational. This result was subsequently tightened by Zudilin,who showed that one of ζ(5), ζ(7), ζ(9), and ζ(11) is irrational.
The Riemann zeta function has an analytic continuation to the entire complexplane except for a simple pole at s = 1. We have another marvelous identity concern-ing its values at negative integers and Bernoulli numbers:
Bn = (−1)nnζ(1 − n)
for natural n.
• T. Rivoal: La fonction zêta de Riemann prend une infinité de valeurs irrationnellesaux entiers impairs, C. R. Acad. Sci. Paris Sér. I Math. 331 (2000), 4, 267–270.
• V.V. Zudilin: One of the numbers ζ(5), ζ(7), ζ(9), ζ(11) is irrational, UspekhiMat. Nauk 56 (2001), 4 (340), 149–150; translation in Russian Math. Surveys 56(2001), 4, 774–776.
Problem 1.38. Let σ(n) denote the sum of divisors of n. Show that σ(n) = 2k if andonly if n is a product of Mersenne primes, i.e., primes of the form 2k −1, for k ∈ Z+.
Solution 1.38. If n = ∏i p
aii , then σ(n) = ∏
i σ (paii ) = ∏
i (1 + pi + · · · + paii ).
Moreover, if each prime factor is a Mersenne number, then pi = 2bi − 1 and ai = 1,which implies that σ(n) = ∏
i 2bi = 2k .Conversely, each prime factor p of n must satisfy 1 + · · · + pa = 2s for some
s ∈ Z. This implies that a = 2q + 1 for q ∈ Z and furthermore, 2s = (1 + p)(1 +p2 + p4 + · · · + p2q). In particular, 1 + p = 2e (thus p is a Mersenne prime) and1 + p2 + · · · + (p2)q = 2n.
If q > 0, then the last equation implies again that q is odd, i.e., q = 2v + 1. Weobtain next (1 + p2)(1 + p4 + · · · + p4v) = 2n, and so 1 + p2 = 2w. In particular,1+p = 2e divides 2w = 1+p2. But 1+p divides 1−p2 and so 1+p2 +1−p2 = 2.
56 5 Number Theory Solutions
Since 1+p ≥ 1+2 = 3, we obtain a contradiction. Therefore, q = 0 and the exponentof the prime is a = 1, as claimed.
Problem 1.39. Find all integer solutions of the equation |pr − qs | = 1, where p, qare primes and r, s ∈ Z \ {0, 1}.Solution 1.39. The only solutions are p = 3, q = 2, r = 2, s = 3 and p = 2, q =3, r = 3, s = 2. Obviously, one prime number should be even, hence 2. Let us assumethat q = 2 and p is odd. Then the equation reads
pr ± 1 = 2s .
If r > 1 is odd, then:1. (pr + 1)/(p + 1) is the odd integer pr−1 − pr−2 − pr−3 − · · · − 1 > 1,
contradiction.2. (pr − 1)/(p − 1) is the odd integer pr−1 + pr−2 + pr−3 + · · · + 1 > 1,
contradiction again.Therefore r = 2t . Further, we have two cases for the equation above:1. 2s = (
pt)2 + 1 = (2n+ 1)2 + 1 = 4n2 + 4n+ 2. This is impossible since 4
divides 2s since s > 1, while 4 does not divide the right-hand side 4n2 + 4n+ 2.2. 2s = (pt )2 − 1 = 4n2 + 4n = 4n(n+ 1). Now, either n or n+ 1 is odd, which
would lead us to a contradiction as soon as n ≥ 2. Thus n = 1 and the solutions arethose claimed above.
Comments 18 Catalan conjectured in 1844 that 32 − 23 = 1 is the only solutionof the Diophantine equation xm − yn = 1, in the integer unknowns x, y,m, n ≥ 2.This was confirmed in 2002 by P. Mihailescu, who gave a brilliant solution usingcyclotomic fields. Note that it suffices to consider the case in which the exponentsm, n are prime numbers. About one hundred years before Catalan, Euler had provedthat there are no other solutions to the equation |x3 − y2| = 1, using the descentmethod.V.A. Lebesgue showed in 1850 that the equation xm−y2 = 1 has no solutions.More than one century later, in 1961, Chao Ko showed that the equation x2 −yn = 1has no solutions if n ≥ 3. His proof was improved by E. Z.Chein in 1976. More detailsconcerning the history of the conjecture and various partial solutions can be foundin the book of Ribenboim.
• T. Metsänkylä: Catalan’s conjecture: another old Diophantine problem solved,Bull. Amer. Math. Soc. (N.S.) 41(2004), 1, 43–57.
• P. Mihailescu: Primary cyclotomic units and a proof of Catalan’s conjecture, J.Reine Angew. Math. 572 (2004), 167–195.
• P. Ribenboim: Catalan’s Conjecture. Are 8 and 9 the Only Consecutive Powers?Academic Press, Inc., Boston, MA, 1994.
Problem 1.40. Consider an arithmetic progression with ratio between 1 and 2000.Show that the progression does not contain more than 10 consecutive primes.
5 Number Theory Solutions 57
Solution 1.40. Suppose that there are 11 consecutive prime terms in the progression.Let r be the ratio. The respective terms are:
h+ nr, h+ nr + r, h+ nr + 2r, . . . , h+ nr + 10r.
Letp be a prime number less than 12. If r �= 0 (mod p), then there exists some i ≤ 10such that h+ ir ≡ 0 (mod p), and hence h+ ir will not be prime. This shows that r isdivisible by 2, 3, 5, 7, 11 and thus greater than 2000, contradicting our assumptions.
Comments 19 More generally, if the ratio r is less than p1 · · ·pn, then the maximumnumber of consecutive primes is at most pn − 1.
Problem 1.41. Let a1 = 1, an+1 = an + [√an
]. Show that an is a perfect square iff
n is of the form 2k + k − 2.
Solution 1.41. Letnk = 2k+k−2.We will prove by induction first thatank = (2k−1)2
and second that nk−1 < m < nk implies that am is not a perfect square. The assentionis true for k = 1. One proves by induction on i that the formulas
ank+2i = (2k−1 + 1 − i)2 + 2k,
ank+2i+1 = (2k−1 + i)2 + 2k−1 − i,
hold for 0 ≤ i ≤ 2k−1. In particular, if i = 2k−1, we obtain ank+1 = (2k)2. Theother values are not perfect squares, since they are located between two consecutivesquares.
Problem 1.42. Recall that ϕ(n) denotes the Euler totient function (i.e., the numberof natural numbers less than n and prime to n), and that σ(n) is the sum of divisorsof n. Show that n is prime iff ϕ(n) divides n− 1 and n+ 1 divides σ(n).
Solution 1.42. If n is prime, then ϕ(n) = n − 1, σ (n) = n + 1. Conversely, let usassume n ≥ 3. Then ϕ(n) is even and thus n must be odd. If p is an odd prime suchthat pr divides n and r ≥ 2, then pr−1 divides ϕ(n) and thus pr−1 divides n − 1,which is a contradiction, since n and n − 1 are coprime, i.e., gcd(n, n − 1) = 1.Therefore, we have n = p1 · · ·pk , where pi are odd primes.
We have further, ϕ(n) = (p1 − 1) · · · (pk − 1) and σ(n) = (p1 + 1) · · · (pk + 1).This implies that 2k divides both ϕ(n) and σ(n).
If k ≥ 2, then 4 divides ϕ(n) and hence n−1. This implies that gcd(4, n+1) = 2.Since 2k divides σ(n), we find that 2k−1 must divide σ(n)
n+1 . In particular, 2k−1 <σ(n)n
,but
σ(n)
n=
(1 + 1
p1
)· · ·
(1 + 1
pk
)<
(4
3
)k,
which leads to a contradiction when k ≥ 2. Therefore, k = 1 and so n is prime.
Comments 20 An old conjecture due to Lehmer states that if ϕ(n) divides n−1, thenn is prime. P. Hagis Jr. proved that the number of prime divisors τ(n) is either 1 orτ(n) ≥ 298 848 and n > 101937042 if 3 divides n, and τ(n) ≥ 1991 and n > 108171
in general. The reader may consult:
58 5 Number Theory Solutions
• P. Hagis Jr.: On the equationM · ϕ(n) = n− 1, Nieuw Arch. Wisk. (4) 6 (1988),3, 255–261.
Problem 1.43. A number is called ϕ-subadditive if ϕ(n) ≤ ϕ(k)+ ϕ(n− k) for all ksuch that 1 ≤ k ≤ n−1, andϕ-superadditive if the reverse inequality holds. Prove thatthere are infinitely many ϕ-subadditive numbers and infinitely many ϕ-superadditivenumbers.
Solution 1.43. Let p ≥ 3 prime and 1 < k < p. We have then
ϕ(k)+ ϕ(p − k) ≤ k − 1 + (p − k − 1) = p − 2 < ϕ(p).
Now p − 1 is composite; thus ϕ(p − 1) ≤ p − 1, and hence ϕ(1) + ϕ(p − 1) ≤1 + p − 3 < ϕ(p). Therefore, every prime number p is ϕ-superadditive.
Let r ≥ 2 and set nr = p1 · · ·pr for the product of the first r prime numbers.Consider some natural numberm, such that 1 < m < nr . Let q1, . . . , qs be the primedivisors of m. We have qj ≥ pj and thus s ≤ r , since otherwise we would havem ≥ q1 · · · qs > p1 · · ·pr = nr . Further, one computes
ϕ(m)
m≥
s∏
i=1
(1 − 1
qi
)≥
r∏
i=1
(1 − 1
pi
)= φ(nr)
nr.
This implies that
ϕ(m)+ ϕ(nr −m) ≥ mϕ(nr)
nr+ (nr −m)
ϕ(nr)
nr= ϕ(nr).
Problem 1.44. Find the positive integers N such that for all n ≥ N , we have ϕ(n) ≤ϕ(N).
Solution 1.44. 1. LetN ≥ 5. According to Chebyshev’s theorem, there exists a primenumber p such that
p
[N + 3
3
]< p < 2
[N + 3
2
]− 2 ≤ N + 2.
We haveϕ(p) = p−1 > N+12 ≥ ϕ(N+1) for oddN andϕ(p) = p−1 > N
2 ≥ ϕ(N)
for even N .This means that N ≤ 5, and since ϕ(1) = ϕ(2) = 1, ϕ(3) = ϕ(4) = 2, we have
N ∈ {1, 2, 3, 4}.Problem 1.45. A number n is perfect if σ(n) = 2n, where σ(n) denotes the sum of alldivisors of n. Prove that the even number n is perfect if and only if n = 2p−1(2p−1),where p is a prime number with the property that 2p − 1 is prime.
Solution 1.45. Since n is even, one can write n = 2qm, where q ≥ 1 and m is
odd. One knows that σ(pa) = pa+1−1p−1 if p is prime, and σ(pa1
1 pa22 · · ·pass ) =
σ(pa11 )σ (p
a22 ) · · · σ(pass ) if pj are distinct primes.
5 Number Theory Solutions 59
We have then σ(m) > m, and we write σ(m) = m+ r for some natural numberr ≥ 1. The hypothesis σ(n) = 2n is equivalent then to 2q+1m = (2q+1−1)(m+r) =2q+1m−m+ (2q+1 − 1)r . Thus m = (2q+1 − 1)r . We have found therefore that rdivides m and r < m. But σ(m) = m+ r is the sum of all divisors of m, and thus ris the sum of all divisors of m that are strictly smaller than m. Since r itself is such adivisor, it follows that it cannot be any other divisor ofm smaller than r and so r = 1.In particular, this means that m is prime.
This means that σ(m) = m + 1 = 2q+1 − 1 and 2q+1m = (2q+1 − 1)(m + 1).Since m is odd and relatively prime to m+ 1, it follows that m = 2q+1 − 1. Further,m has to be prime and so p = q + 1 is a prime number. In fact, if p is composite, sayp = ab, then 2p − 1 = (2a − 1)(1 + 2a + · · · + 2a(b−1)). This proves the claim.
Comments 21 Perfect numbers were known in antiquity. Euclid stated in Book IXof his Elements that a number of the form 2p−1(2p − 1) for which the odd factor isprime is a perfect number, and noticed that 6, 28, 496, and 8128 are perfect. Descartesclaimed in 1638, in a letter to Mersenne, that every even perfect number should beof this form, and this was proved later by Euler to be true. Euler also consideredthe problem of the existence of odd perfect numbers. Although numbers up to 10300
have been checked, no odd perfect numbers have been found to this day, supportingthe conjecture that there are no odd perfect numbers. This is one of the oldest openproblems in mathematics, along with the congruent number problem.
Dickson proved in 1913 that for each k, there are at most finitely many odd perfectnumbers with k distinct prime factors, and the best estimate known (due to PaceP. Nielsen) is that such a number is smaller than 24k . Here are some properties thatan odd perfect number must satisfy:
• P. Hagis Jr. and, independently, J.E.Z. Chein proved that it has at least 8 distinctprime factors (at least 11 if it is not divisible by 3);
• H.G. Hare showed that it has at least 47 prime factors in total, including repeti-tions;
• P. Jenkins proved that it has at least one prime factor greater than 107, andD.E. Iannucci showed that it has two prime factors greater than 104, and threeprime factors greater than 100;
• it is divisible by the power of a prime number that is greater than 1020;• it is a perfect square multiplied by an odd power of a single prime.
Primes of the form 2p − 1 (so p is also prime) are called Mersenne primes. It isstill unknown whether there exist infinitely many Mersenne primes. The first few suchprimes correspond to p = 2, 3, 5, 7, 13, 17, 19, 31, 61, 89. As of October 2006, only44 Mersenne primes were known; the largest known prime number, 232 582 657 − 1, isa Mersenne prime having 9 808 358 digits. It was discovered in September 2006 byCurtis Cooper and Steve Boone.
• R.K. Guy: Unsolved Problems in Number Theory, section B1, 2nd ed. New York,Springer-Verlag, 44–45, 1994.
60 5 Number Theory Solutions
• P. Hagis, Jr.: An outline of a proof that every odd perfect number has at least eightprime factors, Math. Comput. 34 (1980), 1027–1032.
• D.E. Iannucci: The second largest prime divisor of an odd perfect number exceedsten thousand, Math. Comput. 68 (1999), 1749–1760.
• D.E. Iannucci: The third largest prime divisor of an odd perfect number exceedsone hundred, Math. Comput. 69 (2000), 867–879.
• P. Jenkins: Odd perfect numbers have a factor that exceeds 107, Math.Comput.72 (2003), 1549–1554.
• P.P. Nielsen: An upper bound for odd perfect numbers, Integers 3(2003), A14,(electronic).
Problem 1.46. A number n is superperfect if σ(σ(n)) = 2n, where σ(k) is the sumof all divisors of k. Prove that the even number n is superperfect if and only if n = 2r ,where r is an integer such that 2r+1 − 1 is prime.
Solution 1.46. If n = 2r and 2r+1 − 1 is prime, then σ(σ(n)) = σ(2r+1 − 1) =2r+1 = 2n. Conversely, let us consider n = 2rq, where q is odd and r ≥ 1. Then wehave the identity
2r+1q = 2n = σ(σ(n)) = σ((2r+1 − 1)σ (q)).
Moreover, if q > 1, then the number (2r+1−1)σ (q) has at least three distinct divisors,namely (2r+1 − 1)σ (q), σ (q), and 2r+1 − 1. Thus
σ((2r+1 − 1)σ (q)) ≥ 2r+1σ(q)+ 2r+1 − 1 > 2r+1q,
which is a contradiction. Therefore q = 1 and 2r+1 = σ(2r+1 − 1); thus 2r+1 − 1 isprime, as claimed.
Comments 22 It is still unknown whether there exist odd superperfect numbers. Ifone exists, it must be a square and greater than 7 · 1024, as was proved by J.L.Hunsucker and C. Pomerance. Earlier, in 1967, D. Suryanarayana proved that theredo not exist odd superperfect numbers that are powers of a prime.
• J.L. Hunsucker and C. Pomerance: There are no odd super perfect numbers lessthan 7 · 1024, Indian J. Math. 17 (1975), 3, 107–120.
Problem 1.47. Ifa, b are rational numbers satisfying tan aπ = b, thenb ∈ {−1, 0, 1}.Solution 1.47. Write a = m
n, where gcd(m, n) = 1. De Moivre’s identities read
(cos aπ+i sin aπ)n = cosmπ+i sinmπ, (cos aπ−i sin aπ)n = cosmπ−i sinmπ.
Therefore(
1 + ib
1 − ib
)n=
(1 + i tan aπ
1 − i tan aπ
)n=
(cos aπ + i sin aπ
cos aπ − i sin aπ
)n= 1.
5 Number Theory Solutions 61
Consider the polynomial P = xn − 1 ∈ Z[i][x]. Since P is a monic polynomial andZ[i] is a factorial ring, any root of P that lies in Q[i] actually belongs to the ring ofintegers Z[i]. This means that
1 + ib
1 − ib= 1 − b2
1 + b2 + i2b
1 + b2 ∈ Z[i]
and hence 1−b2
1+b2 ∈ Z, 2b1+b2 ∈ Z. Now it is immediate that b ∈ {−1, 0, 1}.
Problem 1.48. Let An = run + svn, n ∈ Z+, where r, s, u, v are integers, u �= ±v,and Pn be the set of prime divisors of An. Then P = ⋃∞
n=0 Pn is infinite.
Solution 1.48. We suppose that rsuv �= 0. We can assume, without loss of generality,that gcd(ru, sv) = 1, and so the prime numbers in P do not divide rsuv. Let ussuppose that P is finite, say {p1, p2, . . . , pm}.
We have ui �= vi , for any i �= 0, because u �= ±v. Then there exists some asuch that ui �≡ vi (mod pak ) for all 1 ≤ i ≤ m + 1, 1 ≤ k ≤ m. Moreover, if1 ≤ i < j ≤ m+ 1, then
Aj − vj−iAi = ruj(uj−i − vj−i
),
and therefore we cannot have Ai ≡ Aj (mod pak ), for all k ≤ m. Thus there existst within the range 1 ≤ t ≤ m + 1 such that At �≡ 0 (mod pak ) for 1 ≤ k ≤ m. Wechoose now some b such that ub ≡ vb (mod pak ) for all 1 ≤ k ≤ m. We can take, forinstance, b = ϕ(ca), where ϕ is Euler’s totient function and c = p1 · · ·pm. We nowchoose n large enough that At+nb > ca . Since we have At+nb ≡ At �≡ 0 (mod pak )for all 1 ≤ k ≤ m, we obtain thatAt+nb must also have a prime divisor different fromp1, . . . , pm.
Comments 23 This problem is a particular case of a result of G. Pólya, see also:
• G. Pólya: Arithmetische Eigenschaften der Reihenentwiclungen rationalen Func-tionen, J. Reine Angew. Math. 151 (1921), 1–31.
Problem 1.49. Solve in natural numbers the equation
x2 + y2 + z2 = 2xyz.
Solution 1.49. If x, y, z are odd, then the left-hand side is odd and therefore nonzero.Thus 2xyz ≡ 0 (mod 4). Each square is congruent to either 0 or 1 modulo 4, andthus x, y, z must be even. Set x = 2x1, y = 2y1, z = 2z1. The equation becomesx2
1 + y21 + z2
1 = 4x1y1z1. The same argument shows that x1, y1, z1 are even. Byinduction, we find a sequence of triples with xn = 2xn+1, yn = 2yn+1, zn = 2zn+1,satisfying
x2n+1 + y2
n+1 + z2n+1 = 2n+2xn+1yn+1zn+1.
Moreover, if n ≥ max(|x|, |y|, |z|), then |xn| = | 12n+1 x| < 1 and similarly for the
other unknowns. Since xn is an integer, we obtain xn = 0 and thus x = y = z = 0.
62 5 Number Theory Solutions
Problem 1.50. Find the greatest common divisor of the following numbers:C12n, C
32n,
C52n, . . . , C
2n−12n .
Solution 1.50. We have the identity C12n + · · · + C2n−1
2n = 22n−1, which can beobtained by writing the binomial expansion of (x + 1)n and evaluating it at x = 1and x = −1. Therefore, the greatest common divisor g should be of the form 2p, forsome natural number p.
If n = 2kq, where q is odd, then C12n = 2k+1q and hence g ≤ 2k+1. But now
we can compute Cp2n = Cp
2k+1q= 2k+1q
pCp−12k+1q−1
. Since p is odd and the binomial
coefficient is an integer, it follows that 2k+1 divides Cp2n for all odd p. This provesthat the number we are seeking is g = 2k+1.
Problem 1.51. Prove that if Sk = ∑ni=1 k
gcd(i,n), then Sk ≡ 0 (mod n).
Solution 1.51. 1.∑ni=1 k
gcd(i,n) = ∑d/n ϕ
(nd
)kd . If gcd(m, n) = 1, then we have
the following identity:
mn∑
i=1
kgcd(i,mn) =∑
d/m
ϕ(md
)∑
τ/n
ϕ(nτ
)(kd)τ .
Thus it suffices to prove the claim for n = pα , where p is prime.The case α = 1 or p divides k is trivial. Let α > 1 and gcd(p, k) = 1. We will
prove the claim by induction on α, as follows:
∑ϕ
(pα
pi
)kp
i = kpα + ϕ(p)kp
α−1 + ϕ(p2)kpα−2 + · · · + ϕ(pα)k
≡ kpα−1 + (p − 1)kp
α−1 + · · · + ϕ(pα)k
≡ p(kpα−1 + · · · + ϕ(pα−1))k
≡ p
α−1∑
i=0
ϕ
(pα−1
pi
)kp
i
(mod pα).
This proves the induction step.2. One observes that 1
nSk counts the number of circular permutations of length k
on n elements. Hence, Sk ≡ 0 (mod n).
Problem 1.52. Prove that for any integer n such that 4 ≤ n ≤ 1000, the equation
1
x+ 1
y+ 1
z= 4
n
has solutions in natural numbers.
Solution 1.52. It suffices to consider the case n prime, since a solution for n yields asolution for a multiple of n. Let a, b, c, d ∈ Z+. If
5 Number Theory Solutions 63
an+ b + c = 4abcd,
then (x, y, z) = (bcd, abdn, acdn) is a solution. In particular, take a = 2, b =1, c = 1. Then, if n = 4d − 1, we obtain a solution as above. This works actually forany n ≡ −1 (mod 4). Taking a = 1, b = 1, c = 1, we solve the case n ≡ 2 (mod 4).If n ≡ 0 (mod 4), then we have the solutions x = y = z = 3n/4. It remains to checkthe case n ≡ 1 (mod 4).
Further, take a = 1, b = 1, c = 2, and thus we solve the case n ≡ 3 (mod 8).The only case remaining is then n ≡ 1 (mod 8).
If n = 3, we have the solutions (x, y, z) = (3, 2, 2). Write
an+ b = c(4abd − 1) = cq, where q ≡ −1 (mod 4ab).
Take q = 3, a = b = 1; then n ≡ −1 (mod 3) is solved. There remains the casen ≡ 1 (mod 3), which we will suppose from now on.
If n = 7, then we have the solution (x, y, z) = (2, 28, 28). Next take q = 7 andab ∈ {1, 2}. This leads to solutions for n ≡ −1,−2,−4 (mod 7). Thus it suffices toconsider n ≡ 1, 2, 4 (mod 7).
Take q = 15 and a = 2, b = 1 or a = 1, b = 2. This solves the equation forn ≡ −2,−8 (mod 15). Since n ≡ 1 (mod 3), we have n ≡ −2,−3 (mod 5). Hence,we may suppose that n ≡ 1, 4 (mod 5).
By asking that n satisfy all these congruences, it follows that the problem is solvedfor all n such that
n �≡ 1, 121, 169, 289, 361, 529 (mod 840).
Since the first prime number not satisfying this condition is 1009, the claim follows.
Comments 24 Erdos and Straus conjectured that the equation 1x
+ 1y
+ 1z
= 4n
hassolutions for every n ≥ 4. Mordell proved that this conjecture is true, except possiblyfor the primes in six particular residue classes modulo 840. Vaughan showed that asufficient condition for the more general conjecture in which 4
nis replaced by m
nis
the existence of positive integers a, b for which an + b has a divisor congruent to−1 (mod mab). The conjecture was verified by I. Kotsireas, via computer search, forn ≤ 1010.
Schinzel conjectured further that the equation
m
n= 1
x1+ · · · + 1
xk
has positive integer solutions xi if m > k ≥ 3; actually, the truth of the conjecturefor m > k = 3 implies its validity for every m > k ≥ 3. See also the book of R.K.Guy which contains more partial results on this topic.
• P. Erdos: On a Diophantine equation (Hungarian, English summary) MatematikaiLapok 1 (1950), 192–210.
• R.K. Guy: Unsolved Problems in Number Theory, Section D11, Springer, NewYork, 1981.
64 5 Number Theory Solutions
• L.J. Mordell: Diophantine Equations, Pure and Applied Mathematics, Vol. 30Academic Press, London-New York 1969.
• A. Schinzel: Sur quelques propriétés des nombres 3/n et 4/n, où n est un nombreimpair, Mathesis 65 (1956), 219–222.
• W. Sierpinski: Sur les décompositions de nombres rationnels en fractions pri-maires, Mathesis 65 (1956), 16–32.
Problem 1.53. For a natural number n, we set S(n) for the set of integers that can bewritten in the form 1 + g + · · · + gn−1 for some g ∈ Z+, g ≥ 2.
1. Prove that S(3) ∩ S(4) = ∅.2. Find S(3) ∩ S(5).
Solution 1.53. 1. Let g, h ≥ 2 be integers satisfying the equation
1 + g + g2 = 1 + h+ h2 + h3
and therefore h3 = (g − h)(g + h + 1). Any prime divisor p of h divides g − h org + h+ 1.
In the first case, p divides g, and in the second case, p divides g+1. Furthermore,if pm divides h then either p3m divides g − h or p3m divides g + h + 1, becausegcd(g, g + 1) = 1. Thus we can write h = h1h2, where gcd(h1, h2) = 1, so that
h31 = g − h, h3
2 = g + h+ 1.
In particular, we have 2h1h2 = 2h = h32 − h3
1 − 1.We will show that this Diophantine equation does not have any solutions h1, h2 ≥
0. Let fh1(h2) = h32 −h3
1 − 2h1h2 − 1. First, we have fh1(h2) ≤ −1 if 0 ≤ h2 ≤ h1.Moreover, if h2 > h1, then fh1 is increasing and so fh1 does not have any naturalzeros, since
fh1(h1 + 1) = h21 + h1 > 0.
This proves that S(3) ∩ S(4) = ∅.2. Let f (x) = 1 + x + x2 and g(x) = 1 + x + x2 + x3 + x4. Then f and g are
strictly increasing functions on Z+. Moreover, we have the following inequalities:
[f (4n2 + 5n+ 1) < g(2n+ 1) < f (4n2 + 9n+ 4) < g(2n+ 2)
< f (4n2 + 9n+ 5) < f (4(n+ 1)2 + 5(n+ 1)+ 1), for all n ≥ 1.
This implies that g(n) �= f (k) if n ≥ 3, k ≥ 1 are natural numbers. Finally, ifn ∈ {1, 2}, then f (1) = 3 < g(1) = 5 < 7 = f (2), and g(2) = 31 = f (5).Therefore S(3) ∩ S(5) = {31}.Problem 1.54. If k ≥ 202 and n ≥ 2k, then prove that Ckn > nπ(k), where π(k)denotes the number of prime numbers smaller than k.
5 Number Theory Solutions 65
Solution 1.54. 1. Let us show first that Ck2k >4kk
if k ≥ 4. If k = 4, then Ck2k = 70 ≥64 = 4k
k. We will use next a recurrence on k. We have
Ck+12k+2 = (2k + 2)(2k + 1)
(k + 1)2Ck2k >
2(2k + 1)
k + 1· 4k
k>
4k · 4k
k · (k + 1)= 4k+1
k + 1.
2. We will use the well-known estimate
π(k) <k
log k − 32
, for k > 5.
The inequality4k
k> (2k)k/(log k−3/2)
is equivalent to k log 4 − log k > klog k+log 2log k−3/2 , which is true as soon as k ≥ 1414.
Moreover,
Ck2k >4k
k> 2k
k/(log k−3/2)> (2k)π(k), for k ≥ 1414.
By direct computer verification, one has Ck2k > (2k)π(h) for 202 ≤ h ≤ 1413.3. Now if s ≥ 1, we have
k > π(k) ≥ 1 + π(k)− 1
s= π(k)+ s − 1
s.
The product of these inequalities for 1 ≤ s ≤ r gives us kr > Crπ(k)+r−1. Then
Ckn+1 =(
1 − k
n+ 1
)−1
Ckn >
(1 − k
n+ 1
)−1
nπ(k) = nπ(k) ·∞∑
r≥0
kr
(n+ 1)r
> nπ(k) ·∞∑
r≥0
Crπ(k)+r−1
(n+ 1)r= nπ(k)
(1 − 1
n+ 1
)−π(k)= (n+ 1)π(k).
Problem 1.55. Let Sm(n) be the sum of the inverses of the integers smaller than mand relatively prime to n. If m > n ≥ 2, then show that Sm(n) is not an integer.
Solution 1.55. Let us write
Sm(n) = 1 + 1
a1+ · · · + 1
as,
where 1 < a1 < a2 < · · · < as < m and gcd(ai, n) = 1. Observe first that a = a1 isprime, since otherwise, any proper divisor of a would be smaller and still relativelyprime to n. Let k be the largest integer such that ak ≤ m. Then there exists somei ≤ s such that ak = ai , because ak is also a number relatively prime to n.
Moreover, if ak divides aj for some j , then j = i. In fact, let us assume thataj = cak for some integer c > 1. If c < a, then by hypothesis, gcd(c, n) > 1, and
66 5 Number Theory Solutions
hence aj would not be relatively prime to n. If c ≥ a, then m > aj = cak ≥ ak+1,which would contradict the choice of k, supposed to be maximal with the propertythat ak ≤ m.
Let us suppose now that Sm(n) is an integer, and set L = lcm(a1, . . . , as). Oneknows from above that ak divides L, while ak+1 does not divide L.
We have also
LSm(n)−s∑
j=1,j �=i
L
aj= L
ai.
If Sm(n) were an integer, then the left-hand side would be a multiple of a. On theother hand, the term on the right-hand side is L
ai= L
ak, which cannot be a multiple of
a, by the choice of k. This contradiction proves that Sm(n) is not an integer.
Problem 1.56. Solve in integers the following equations:
1. x4 + y4 = z2;2. y3 + 4y = z2;3. x4 = y4 + z2.
Solution 1.56. 1. Suppose that there exists a nontrivial solution for which x or y isnonzero and hence a solution with x, y, z > 0. Consider the solution for which z isminimal among all possible solutions. We may suppose that gcd(x, y, z) = 1. Thenthe triple (x2, y2, z) is a primitive Pythagorean triple and so gcd(x, y) = gcd(y, z) =gcd(x, z) = 1. We can write x2 = m2 − n2, y2 = 2mn, z = m2 + n2, where m, nare integers. Since the solutions are primitive, it follows that x and z must be odd. Inparticular, m �≡ n (mod 2), since otherwise, x would be even. We have two cases:
(i). If m is even and n is odd, then x2 = m2 − n2 ≡ −1 (mod 4), which isimpossible, since any square is congruent to either 0 or 1 modulo 4.
(ii). If m is odd and n is even, then we have y2 = 2mn and gcd(m, n) = 1. Thusm = m2
1 and n = 2n21. Moreover, x2 = m4
1 − 4n41, and so
(m21 − x)(m2
1 + x) = 4n41.
If d divides bothm21 −x andm2
1 +x, then d divides 2m and n and so d = 2 or d = 1.But m2
1 − x and m21 + x are congruent modulo 2, and they cannot be both odd, since
their product is even. Thus m21 − x = 2a4 and m2
1 + x = 2b4. In particular, we findthat
a4 + b4 = m21,
where a, b are nonzero and m1 < m21 = m < z. Thus we have obtained a solution
with z smaller than the initial one, which is a contradiction. Therefore there are nonontrivial solutions.
2. We have (y3 + 4y)2 = y6 + 8y4 + 16y2 = z4. Also, (y3 − 4y)2 = y6 +16y2 − 8y4. Adding these equations, we obtain (2y)4 = z4 − (y3 − 4y)2. It is knownthat the equation u4 − v4 = w2 does not have nontrivial integer solutions (see theprevious equation). Therefore either 2y = 0 and z4 = (y3 − 4y)2 = 0, which leadsto y = z = 0, or else z = 2y, y3 − 4y = 0, which implies y = 2, z = 4.
5 Number Theory Solutions 67
3. Consider a positive solution (x, y, z) with gcd(x, y) = 1 and minimal x.Then (y2, z, x2) is a primitive Pythagorean triple, and hence we have two possiblesituations:
(i). Either y2 = m2 −n2, x2 = m2 +n2, z = 2mn, wherem > n and gcd(m, n) =1. Then
a4 = b4 + (xy)2
and a < x, contradicting the fact that x was chosen minimal.(ii). Or z = m2 −n2, x2 = m2 +n2, y2 = 2mn, wherem > n and gcd(m, n) = 1.
Then (m, n, x) is also a primitive Pythagorean triple and thusm is odd and n is even.Since y2 = 2mn and n is odd, we can write m = 2a2, n = b2, so that y = 2ab.Using the fact that (m, n, x) is a primitive Pythagorean triple, where m is even and nis odd, we derive thatm = 2rs, n = r2 − s2 = b2, with gcd(r, s) = 1. Thus a2 = rs,which yields r = u2, s = v2, and thus, expressing n as above, we obtain:
u4 = v4 + b2
with u < r < m < x, contradicting the minimality of x. Thus there are no nontrivialsolutions.
Comments 25 In particular, the Fermat equation x4 + y4 = z4 has no nontrivialsolutions. The same method, called infinite descent, can be used successfully to provethat the equations x4 + dy4 = z2 have no solutions, when d is prime and d ≡7, 11 (mod 16), or d
2 is prime and d2 ≡ ±3 (mod 8), or −d is prime and −d ≡
±3, 11 (mod 16), or d4 is prime and d
4 ≡ ±3, 11 (mod 8). For more informationabout Diophantine equations see
• L.J. Mordell: Diophantine Equations, Academic Press, Pure and Applied Math.vol. 30, London, New York, 1969.
Problem 1.57. Show that the equation
x3 + y3 = z3 + w3
has infinitely many integer solutions. Prove that 1 can be written in infinitely manyways as a sum of three cubes.
Solution 1.57. 1. We make the change of variables x+y = X, x−y = Y , z+w = Z,z− w = W . The equation now reads
X(X2 + 3Y 2
)= Z
(Z2 + 3W 2
).
We use the identity(a2 + 3b2
) (A2 + 3B2
)= (aA+ 3bB)2 + 3(bA− aB)2
and derive
68 5 Number Theory Solutions
X(X2 + 3Y 2
)2 = Z((XZ + 3YW)2 + 3(YZ −XW)2
).
Set
p = XZ + 3YW
X2 + 3Y 2 , q = YZ −XW
X2 + 3Y 2 ,
and thus XZ
= p2 + 3q2. Choose Z = 1, for simplicity. We have then
pX + 3qY = 1, pY − qX = W,
from which one can obtain the values of Y and W . Finally, we obtain
x = 1 − (p − 3q)(p2 + 3q2
), y = −1 + (p + 3q)
(p2 + 3q2
),
z = p + 3q −(p2 + 3q2
)2, w = −p + 3q +
(p2 + 3q2
)2.
These are all rational solutions of our equation. If p, q ∈ Z, then we obtain infinitelymany integer solutions. However, it is not clear that these formulas will provide allinteger solutions of our equations.
2. Set p = 3q, t = 2q. We obtain
1 =(
1 − 9t3)3 +
(3t − 9t4
)3 +(
9t4)3.
Comments 26 It is unknown whether 3 can be written in infinitely many ways as asum of three cubes.
Problem 1.58. Prove that the equation y2 = Dx4 + 1 has no integer solution, exceptx = 0, if D �≡ 0,−1, 3, 8 (mod 16) and there is no factorization D = pq, wherep > 1 is odd, gcd(p, q) = 1, and either p ≡ ±1 (mod 16), p ≡ q ± 1 (mod 16), orp ≡ 4q ± 1 (mod 16).
Solution 1.58. Let x, y > 0 be a nontrivial solution with minimal x. If x is odd, thenDx4 +1 �≡ 0, 1, 4, 9 (mod 16), and thus it is not a quadratic residue modulo 16. Thusx should be even and y odd. Further, we have
y + 1 = 2pa4, y − 1 = 8qb4, x = 2ab
ory − 1 = 2pa4, y + 1 = 8qb4, x = 2ab,
where D = pq, gcd(p, q) = 1, and a is odd.If p > 1, then
pa4 − 4qb4 = ±1.
If b is even, then p ≡ ±1 (mod 16), contradiction. If b is odd, then p ≡ 4q ±1 (mod 16), again false.
Thus p = 1 anda4 − 4Db4 = ±1.
5 Number Theory Solutions 69
The equation a4 − 4Db4 = −1 has no solution modulo 4; thus a4 − 1 = 4Db4. ThenD = rs, b = cd , gcd(r, s) = 1, and
a2 + 1 = 2rc4, a2 − 1 = 2sd4.
Thus c, r are odd, and we have
rc4 − sd4 = 1.
If r = 1, then c4 −Dd4 = 1 and x = 2acd, so there exists a solution of our equationwith d ≤ x/2, contradicting the minimality of x.
If r > 1, then d cannot be even because r �≡ ±1 (mod 16). But if d is odd, thenr ≡ s + 1 (mod 16), contradicting our assumptions.
Problem 1.59. Find all numbers that are simultaneously triangular, perfect squares,and pentagonal numbers.
Solution 1.59. We have the Diophantine equations
n = p2 = 1
2q(q + 1) = 1
2r(3r − 1).
We derive that
(2q + 1)2 = 1 + 8p2and (6r − 1)2 = 1 + 24p2.
In particular, we obtain
(6r − 1)2 = (4p)2 + (2q + 1)2.
Since gcd(4p, 2q+1) = 1, the triple (4p, 2q+1, 6r−1) is a primitive Pythagoreantriple. This means that the integer solutions are determined as functions of two integerparameters m, n ∈ Z by means of the formulas
4p = 2mn, 2q + 1 = m2 + n2, 6r − 1 = m2 + n2.
Substituting in the previous equation, we obtain them4 − 4m2n2 +n4 = 1 constrainton the parameters, which can also be written as (m2 − 2n2)2 = 3n4 + 1.
We now use a theorem that says that the Ljunggren equation
x2 = Dy4 + 1
has at most two positive integer solutions if D > 0 is not a perfect square.In our case, the equation x2 = 3y4 + 1 has the obvious solutions (7, 2) and (2, 1)
and thus these are all the solutions in natural numbers. Therefore there is only onesolution (m, n) = (2, 1).
70 5 Number Theory Solutions
Comments 27 The proof above is based on a deep result concerning the equationx2 = Dy4 + 1, proved by Ljunggren in 1942. Ljunggren also gave an algorithm thatcomputes the nontrivial solution when it exists. The previous problem is a particularcase in which a short elementary proof (due to L. Mordell) is available.
Ljunggren extended his methods to deal with the more-delicate equation x2 = Dy4 −1. He proved that this Diophantine equation has at most two positive solutions forD > 0, not a perfect square. In particular, x2 = 2y4 − 1 has only the solutions (1, 1)and (13, 239), which have been known for two centuries. Recently, Chen and Voutierproved that there is at most one solution in positive integers when D ≥ 3.
• W. Ljunggren: Einige Eigenschaften der Einheiten reeller quadratischer und rein-biquadratischer Zahlkörper mit Anwendung auf die Lösung einer Klasse unbe-stimmter Gleichungen vierten Grades, Skr. Norske Vid. Akad. Oslo 1936, 12,1–73.
• W. Ljunggren: Über die Gleichung x4 − Dy2 = 1, Arch. Math. Naturvid. 45(1942) 5, 61–70.
• W. Ljunggren: Zur Theorie der Gleichung x2 +1 = Dy4, Avh. Norske Vid. Akad.Oslo. I. 1942.
• J.H. Chen and P.Voutier: Complete solution of the Diophantine equationX2+1 =dY 4 and a related family of quartic Thue equations, J. Number Theory 62 (1997),1, 71–99.
Problem 1.60. Find all inscribable integer-sided quadrilaterals whose areas equaltheir perimeters.
Solution 1.60. We have 16S2 = ∏cyclic(a + b + c − d), and so the problem is
equivalent to the Diophantine equation
16(a + b + c + d)2 =∏
cyclic
(a + b + c − d).
Assume that 0 < d ≤ c ≤ b < a ≤ b+ c+ d, for the sake of simplicity. If we writex = −a + b + c + d , y = a − b + c + d, z = a + b − c + d, t = a + b + c − d ,then the above equation becomes
(x + y + z+ t)2 = xyzt.
If this equation has a real solution, then the quadratic equation (X + y + z + x)2 −Xxyz = 0 has two real roots. The product of these real roots is (x + y + z)2. Sincethe solution t satisfies 0 < t < x + y + z, it follows that t is the smallest of the tworoots, and thus it is determined by the formula
2t = xyz− 2(x + y + z)− √xyz(xyz− 4x − 4y − 4z)
with the obvious constraint xyz > 4(x + y + z).Therefore, we need that λ = x+y+z
xyz< 1
4 . Now, 3t ≥ x + y + z, since t is the
longest side, and thus 83λ+ √
1 − 4λ ≤ 1, implying that λ ≥ 316 .
5 Number Theory Solutions 71
We have obtained the inequalities 14 > λ ≥ 3
16 , which immediately imply that5 ≤ xy ≤ 16.
Since λ < 14 , we get z > 4(x + y)/(xy − 4).
On the other hand, we assumed that z ≤ t , which is equivalent to
(xy − 4)z2 − 4(x + y)z− (x + y)2 ≤ 0,
and therefore z ≤ (x + y)/(√xy − 2). Consequently,
4(x + y)
xy − 4≤ z ≤ x + y√
xy − 2.
One can list all integer values of x, y, z, t satisfying the inequalities above andthat are solutions to the Diophantine equation. This leads to
(x, y, z, t) ∈ {(1, 9, 10, 10), (2, 5, 5, 8), (3, 3, 6, 6), (4, 4, 4, 4)}and their permutations.
Problem 1.61. Every rational number a can be written as a sum of the cubes of threerational numbers. Moreover, if a > 0, then the three cubes can be chosen to bepositive.
Solution 1.61. Assume that we know how to write a = x3 + y3 + z3 and let usintroduce u = x+y+z and v = x+y. Then a = (x+y+z)3−3(x+y)(y+z)(z+x),and we derive easily that a = u3 − 3v(u− x)(u− v + x).
We will require furthermore that u and v satisfy the additional constraint u3 =3v(u2 − x2). Then a = 3v2(u− x).
We introduce the new variable w = xu
∈ Q, so that u = 3v(1 − w2). Thereforea = 3v2(u− x) = 3v2u(1 −w) = 9v3(1 −w)2(1 +w) = 27v3(1 −w)3 · 1
3 · 1+w1−w .
Set m = 3v(1 − w) and introduce it above in order to obtain a = m3
31+w1−w .
Now we reverse the line of thought by letting m ∈ Q be an arbitrary parameter.The previous formulas yield w = 3a−m3
3a+m3 and also the following solutions:
v = 3a +m3
6m2 , u = 6am
3a +m3 .
This yields the rational solutions x = wu, y = v − wu, z = u− v.Let us prove the positivity part. We have that x = uw ≥ 0 is equivalent to
0 < m <3√
3a. But let us assume for a moment that m = 3√
3a. Then y = 3√3a3 , z =
2 3√3a3 , x = 0. Regarding the solutions y, z as functions y = y(m), z = z(m) of
the parameter m, we immediately see that these are continuous functions, which arestrictly positive at m = 3
√3a. In particular, there exists a small neighborhood W of
3√
3a such that y(m), z(m) > 0. This means that for m ∈ Q+ ∩ (0, 3√
3a) ∩W, wehave x, y, z ∈ Q+, as claimed.
72 5 Number Theory Solutions
Problem 1.62. 1. Every prime number of the form 4m + 1 can be written as thesum of two perfect squares.
2. Every natural number is the sum of four perfect squares.
Solution 1.62. 1. Let p be such a prime number. The congruence z2 +1 ≡ 0 (mod p)has at least one solution, because −1 is a quadratic residue when p is of the form4m+ 1. This follows from the more general Euler criterion computing the Legendre
symbol(ap
)as follows:
(a
p
)= a
p−12 (mod p)
for arbitrary prime p. Recall that the Legendre symbol(ab
)equals 1 if a is a quadratic
residue modulo b (i.e., there exists some integer n such that n2 ≡ a(mod b)), and −1otherwise.
Let then m ∈ Z+, m �= 0, and z ∈ Z be such that
mp = z2 + 1.
We can suppose that −p2 < z <
p2 , by using a translation multiple of p. Then
m = 1p(z2 + 1) < 1
p
(p2
4 + 1)< p. Set x = z, y = 1, so that mp = x2 + y2.
We shall prove that if m > 1, then there exists m′ < m with the same properties.This process stops when m = 1, where we find the wanted sum of two squares.
Let us consider u, v such that −m2 ≤ u, v ≤ m
2 , satisfying
u ≡ x (mod m), v ≡ y (mod m).
Thusu2 + v2 = x2 + y2 ≡ 0 (mod m),
and so there exists some r ∈ Z+ for which
u2 + v2 = mr.
Observe that r �= 0, since otherwise x, y would be multiples of m, and so m would
divide p. Further, r = 1m
(u2 + v2
)< 1
m
(m2
4 + m2
4
)< m. We will use the identity
(a2 + b2
) (A2 + B2
)= (aA+ bB)2 + (aB − bA)2
and multiply the two equalities above, obtaining
m2rp =(x2 + y2
) (u2 + v2
)= (xu+ yv)2 + (xv − yu)2.
Moreover,
xu+ yv ≡ x2 + y2 ≡ 0 (mod m) and xv − yu ≡ xy − yx ≡ 0 (mod m),
5 Number Theory Solutions 73
and thus, dividing the equation by m2, we obtain
rp = a2 + b2,
where r < m. Thus, m′ = r satisfies all required properties. This proves the firstclaim.
2. The identity
(a2 + b2 + c2 + d2)(A2 + B2 + C2 +D2) = (aA+ bB + cC + dD)2
+(aB − bA− cD + dC)2 + (aC + bD − cA− dB)2 + (aD − bC + cB − dA)2
reduces the problem to the case of prime numbers. Moreover, we know that primesof the form 4m+ 1 can be written as a sum of two squares, and 2 = 12 + 12. Thus itsuffices to consider the primes of the form 4m+ 3. The method of proof is the sameas above.
We show first that the congruence x2 + y2 + 1 ≡ 0 (mod p) has solutions.By the Euler criterion, −1 is a quadratic nonresidue modulo p and thus −y2 is aquadratic nonresidue modulo p. Moreover, each quadratic nonresidue is of the form−y2 (mod p). Then we have to find a quadratic nonresidue n and a quadratic residuer such that r + 1 = n (mod p). Consider in the sequence {1, 2, 3, . . . , p − 1, p} thesmallest quadratic nonresidue n. Then n > 1 and n − 1 must be a quadratic residueby the minimality of n, and this yields our pair (r, n).
We write nowmp = x2 + y2 + 12 + 02,
where m ∈ Z+ is nonzero. We choose x, y such that −p2 < x, y <
p2 , using a
translation multiple of p. Then m = 1p
(x2 + y2 + 1
)< 1
p
(p2
4 + p2
4 + 1)< p.
Consider a = x, b = y, c = 1, d = 0. If m > 1, we will find another m′ < m suchthat m′p is still a sum of four squares.
Choose A,B,C,D such that −m2 < A,B,C,D ≤ m
2 and
a ≡ A (mod m), b ≡ B (mod m), c ≡ C (mod m), d ≡ D (mod m).
Then A2 + B2 + C2 +D2 ≡ 0 (mod m) and thus there exists r ∈ Z+ such that
mr = A2 + B2 + C2 +D2.
Furthermore, r �= 0, since otherwise,A = B = C = D = 0, and thus a, b, c, d wouldbe multiples ofm, and somwould dividep. Moreover, r = 1
m(A2+B2+C2+D2) ≤
m, with equality if and only if A = B = C = D = m2 , m even. The equality would
imply that a2 = m2
4 (mod p) and thus pm ≡ 0 (mod m2). Thus m divides p, whichis false. Thus r < m. Further, multiplying the two identities, we obtain
m2rp =(a2 + b2 + c2 + d2
) (A2 + B2 + C2 +D2
)= X2 + Y 2 + Z2 +W 2.
The terms X, Y,Z,W are multiples of m, since, for example,
74 5 Number Theory Solutions
X = aA+ bB + cC + dD ≡ a2 + b2 + c2 + d2 ≡ 0 (mod m),
and similarly for Y,Z,W . Therefore, one can divide by m2 and find that
rp = X′2 + Y ′2 + Z′2 +W ′2
with r < m. Then take m′ = r . This proves the claim.
Comments 28 The proof given above was obtained by Lagrange, and all elementaryproofs known until now are variations of the present one. The numbers of the form4m(8k + 7) cannot be represented as sums of three squares. Thus 4 is the minimalnumber with this property. However, all numbers that are not of the form 4m(8k+ 7)are sums of three perfect squares.
Problem 1.63. Every natural number can be written as a sum of at most 53 integersto the fourth power.
Solution 1.63. Consider the identity
6(x2
1 + x22 + x2
3 + x24
)2 =∑
1≤i≤j≤4
(xi − xj )4 +
∑
1≤i<j≤4
(xi + xj )4.
We write n = 6x + y, where 0 ≤ y ≤ 5. According to Lagrange’s theorem (theprevious problem), any natural number is the sum of four perfect squares. Therefore
x = α21 + α2
2 + α23 + α2
4, αi = α2i1
+ α2i2
+ α2i3
+ α2i4.
Then
6x =4∑
i=1
6
⎛
⎝4∑
j=1
α2ij
⎞
⎠
2
=4∑
i=1
∑
1≤j<k≤4
(αij − αik )4 +
4∑
i=1
∑
1≤j<k≤4
(αij + αij + αik )4 =
48∑
i=1
w4i .
Now, any y ∈ {0, . . . , 5} can be written as a sum of at most 5 fourth powers, and thusn can be written as a sum of at most 53 fourth powers.
Comments 29 The problem of determining the minimumG(k) such that any numberis the sum ofG(k)perfect (positive) kth powers (in our case k = 4) is called theWaringproblem. For instance, Lagrange proved in 1770 thatG(2) = 4. Earlier the same year,Edward Waring had conjectured that something similar could be proved for cubes,fourth powers, and so on. He stated, without proof, that it would take the sum of atmost nine cubes or 19 fourth powers to express any positive integer. Waring arrived athis conjectures about cubes and fourth powers by collecting data and looking for pat-terns. The sequence of minimum number of cubes needed to write n ≤ 30 as a sum ofcubes is 1, 2, 3, 4, 5, 6, 7, 1, 2, 3, 4, 5, 6, 7, 8, 2, 3, 4, 5, 6, 7, 8, 9, 2, 3, 4, 1, 2, 3, 4.
5 Number Theory Solutions 75
Fourth powers show similar behavior: 15 can be written as the sum of 15 fourthpowers, 31 requires 16 fourth powers, 47 requires 17, 63 requires 18, and 79 re-quires 19. Among the first 12000 numbers, the only number other than 23 that re-quires nine cubes is 239. Fifteen numbers require a minimum of eight cubes: 15,22, 50, 114, 167, 175, 186, 212, 213, 238, 303, 364, 420, 428, and 454. The list ofnumbers requiring seven cubes is much longer, but it includes no numbers greaterthan 8042.
In 1909, David Hilbert proved that there is some universal constant G(k) such thatevery natural number is the sum of at most G(k) kth powers. However, his proof didnot provide an effectiveG(k). In 1912, A.J. Kempner completed a 1909 partial proofby A. Wieferich to establish, once and for all, that every integer can be expressed asa sum of nine cubes. In 1940, S.S. Pillai showed that every integer can be expressedas a sum of 73 sixth powers.
The easy estimate obtained above, that G(4) ≤ 53, is far from being sharp. Theassertion that 37 fifth powers are sufficient was proved by Chen Jingrun in 1964. Itwasn’t until 1986 that R. Balasubramanian, J.-M. Deshouillers, and F. Dress provedthat G(4) = 19, by showing that it suffices to check the claim for those numbers lessthan 10367.
A related question concerns the number of terms required to express every sufficientlylarge integer as a sum of kth powers. For example, even though every integer can beexpressed as a sum of at most nine cubes, every integer larger than a certain value(probably 8 042) can be written as the sum of at most seven cubes. One conjectures thatevery sufficiently large integer can be expressed as the sum of no more than four cubes.The largest number now known not to be a sum of four cubes is 7 373 170 279 850.Davenport proved that any sufficiently large integer is represented as the sum of 16fourth powers, and Kempner that no integer of the form 16h31 is representable as asum of fewer than 16 integral fourth powers.
In 1995, Irving Kaplansky and Noam D. Elkies proved that any integer can be ex-pressed as the sum of two squares and a cube, when positive and negative integers areallowed. Numerical experiments show evidence for the conjecture that all but finitelymany numbers can be expressed as the sum of a square and two cubes, using onlypositive integers. See also:
• R. Balasubramanian, J.-M. Deshouillers, and F. Dress: Problème de Waring pourles bicarrés. I. Schéma de la solution. C. R. Acad. Sci. Paris Sér. I Math. 303(1986), 4, 85–88.
• H. Davenport: On Waring’s problem for fourth powers, Ann. of Math. (2) 40(1939), 731–747.
• W.C. Jagy and I. Kaplansky: Sums of squares, cubes, and higher powers, Exper-imental Mathematics 4 (1995), 169–173.
• H. Rademacher and O. Toeplitz: The Enjoyment of Mathematics: Selections fromMathematics for the Amateur, New York, Dover, 1990.
76 5 Number Theory Solutions
Problem 1.64. Let G(k) denote the minimal integer n such that any positive integercan be written as the sum of n positive perfect kth powers. Prove that G(k) ≥ 2k +[
3k
2k
]− 2.
Solution 1.64. Let us write 3k = 2kq + r , where 0 < r < 2k and q =[
3k
2k
], and
consider the number A = 2kq − 1. We can write A = (q − 1)2k + (2k − 1)1k , thus
as a sum of 2k +[
3k
2k
]− 2 perfect kth powers. Let us show that we cannot do better.
First, A < 3k , and thus we cannot use 3k among our kth powers. On the other hand,A < 2kq, and thus at most q − 1 terms among our kth powers could be equal to2k . The remaining terms should be 1k , and thus there are at least q − 1 + 2k − 1, asclaimed.
Comments 30 This is the general case of the Waring problem, discussed above. It is
conjectured thatG(k) = 2k +[
3k
2k
]− 2, for any k ≥ 2. Moreover, the conjecture was
proved by L.E. Dickson and (independently) by S.S. Pillai for those k satisfying
r = 3k − 2kq ≤ 2k − q − 2,
where q and r are as above. This inequality was verified for all k ≤ 471 600 000.Moreover, it is known that the inequality holds for all sufficiently large k, but we donot have an effective lower bound.
Problem 1.65. Let R(n, k) be the remainder when n is divided by k and
S(n, k) =k∑
i=1
R(n, i).
1. Prove that limn→∞ S(n, n)
n2 = 1 − π2
12 .2. Consider a sequence of natural numbers (ak) growing to infinity and such that
limk→∞ ak log kk
= 0. Prove that limk→∞ S(kak, k)
k2 = 14 .
Solution 1.65. 1. We calculate that:
S(n+ 1, n+ 1) = S(n, n)− σ(n+ 1)+ 2n+ 1,
where σ(m) denotes the sum of divisors of m. Therefore, we find that S(n, n) =n2 − ∑
j≤n σ (j). Furthermore, use the estimate
∑
j≤nσ (j) = 1
2ζ(2)n2 + O(n log n)
and get the claim.2.We have the identities
5 Number Theory Solutions 77
S(n, k) = nk −k∑
i=1
[ni
]i = nk −
[nk
] k∑
i=1
i −∑
nk<k≤n
⎛
⎝
[nk
]∑
i=1
i
⎞
⎠
in order to obtain the estimate
S(n, k) = nk −[nk
] k(k + 1)
2− n2
2
∞∑
i=[nk
]+1
1
i2+ O(n log k).
This automatically yields the first claim. Further,
S(kak, k) = 1
2
⎛
⎝k2ak − kak − k2a2k
∞∑
i=ak+1
1
i2
⎞
⎠ + O(kak log k),
and we can estimate( ∞∑
i=N+1
1
i2
)
− 1
N + 12
=∞∑
i=N+1
1
i2−
∞∑
i=N+1
(1
N − 12
− 1
N + 12
)
= −1
4
∞∑
i=N+1
1
i4 − i2
4
= O(
1
N3
),
and hence
S(kak, k)
k2 = 1
4
ak
ak + 12
− 1
2k+ O(a−1
k )+ O(ak log k
k
),
which proves the claim.
Comments 31 L. Funar conjectured in 1985 that limn→∞ S(2n, n)n2 = 1 − π2
12 . Thisquestion seems to be still open.
• L. Funar: Problem 6476, Amer. Math. Monthly, 81(1984), 588.
Problem 1.66. Let a1 < a2 < a3 < · · · < an < · · · be a sequence of positiveintegers such that the series
∞∑
i=1
1
ai
converges. Prove that, for any i, there exist infinitely many sets of ai consecutiveintegers that are not divisible by aj for all j > i.
78 5 Number Theory Solutions
Solution 1.66. Fix an arbitrary i. The set Sn = {n − ai, n − ai + 1, . . . , n − 1} iscalled admissible for aj (where j > i) if no element of Sn is divisible by aj . It isobvious that Sn is admissible for aj if aj divides n. Moreover, Sn is not admissiblefor aj if there exists some e, with 1 ≤ e ≤ ai , such that n ≡ e (mod aj ).
Consider ε > 0 sufficiently small and k sufficiently large such that
∑
j>k
a−1j < ε.
Let β = ai+1 · · · ak and Sn with the property that β divides n. Observe first that thesets Sn are admissible for those aj with i ≤ j ≤ k. Now assume that j > k. LetCj (x) denote the set of those c ≤ x for which Scβ is not admissible for aj . Recallthat c ∈ Cj (x) iff there exists e ∈ {1, 2, . . . , ai} such that cβ ≡ e (mod aj ).
If gcd(β, aj ) = d, then d should divide e, and we can write β = dβ ′, e =de′, aj = da′
j . Moreover, the condition above is equivalent to cβ ′ ≡ e′ (mod a′j ).
If aj > βx, then obviously Cj (x) = ∅. Further, for a fixed value of e′, thereare at most x
a′j
+ 1 solutions c for the previous congruence in the given range. Since
e′ can take at most aid
values, we derive that the cardinality of Cj (x) is bounded by
ai
(xaj
+ 1)
.
Therefore, among the multiples of β satisfying n < βx, there are at most
∑
j>k
aix
aj+ ai card{j |aj < βx} = εaix + o(x)
integers n such that Sn is not admissible. In fact, the estimate card{j |aj < βx} = o(x)
is a consequence of the convergence of the series. This asymptotic estimate impliesthat for large x there exist infinitely many values of n that do not belong to any Cj ,so that Sn is admissible with respect to all aj .
Problem 1.67. Denote byCn the claim that there exists a set of n consecutive integerssuch that no two of them are relatively prime. Prove that Cn is true for every n, suchthat 17 ≤ n ≤ 10000.
Solution 1.67. We suppose that both p and 2p + 1 are primes and consider those nfor which 3p+2 ≤ n ≤ p2. Set q for the product
∏pi≤n,pi �=p,2p+1 pi of those prime
numbers smaller than n and distinct from p and 2p+ 1. From the Chinese remaindertheorem, there exists a natural number x satisfying the congruences
x ≡ 0 (mod q), x ≡ −3p − 1 (mod p(2p + 1)).
Now let Sn = {x, x+1, . . . , x+n−1}. Then Sn contains x+1, x+2p+1, . . . , x+kp + 1, . . ., which have p as a common factor. Also, x + p has the factor 2p + 1 incommon with x + 3p + 1 ∈ Sn. All other numbers from Sn have a common divisorwith q, therefore with x. Therefore the set Sn satisfies the claim Cn, and we have tolook for specific values of p.
5 Number Theory Solutions 79
Ifp = 5, thenSn holds for 17 ≤ n ≤ 25.We choose furtherp = 11, 29, 251, 1013,49919, 5008193, and this yields the claim for all n such that 35 ≤ n ≤ (5008193)2.
The case 26 ≤ n ≤ 34 must be treated separately. There one considers x satisfying
x ≡ 0 (mod 2 · 5 · 11 · 17), x ≡ −1 (mod 3), x ≡ −2 (mod 7 · 19),
x ≡ −3 (mod 13), x ≡ −4 (mod 23).
Comments 32 The claim Cn is false for all n ≤ 16, and true for all n ≥ 17, asproved in:
1. S.S. Pillai: Onm consecutive integers, Proc. IndianAcad. Sci., Sect.A. 11 (1940),6–12, 13 (1941), 530–533.
2. A. Brauer: On a property of k consecutive integers, Bull. Amer. Math. Soc. 47(1941), 328–331.
The proof, which is elementary, is based on the estimate
π(2n)− π(n) ≥ 2
[log n
log 2
]+ 2, for n ≥ 75.
Problem 1.68. Prove that a natural number n has more divisors that can be written inthe form 3k + 1, for k ∈ Z, than divisors of the form 3m− 1, for m ∈ Z.
Solution 1.68. Let f (n) (and respectively g(n)) be the number of divisors of the form3k + 1 (and respectively 3m − 1), for k,m ∈ Z. Let n = uv, where gcd(u, v) = 1.By making use of the obvious congruences
(−1)(−1) ≡ 1 (mod 3), (−1)1 ≡ −1 (mod 3),
we obtain the formulas
f (n) = f (u)f (v)+ g(u)g(v), g(n) = g(u)f (v)+ f (u)g(v).
Set further h(n) = f (n)− g(n). The identities above prove that
h(n) = h(u)h(v).
In particular, if n = pα11 · · ·pαrr is the factorization into prime factors of n, then
h(n) =r∏
j=1
h(pαjj
).
Now we haveh(3α) = 1.
If the prime number p ≡ −1 (mod 3), then the divisors of pα of the form 3k + 1 are1, p2, p4, . . . and the ones of the form 3k − 1 are p, p3, . . . . Therefore
80 5 Number Theory Solutions
h(pα) ={
1, if α is even0, otherwise.
If the prime number p satisfies p ≡ 1 (mod 3), then all divisors of pα are of the form3k + 1, and therefore h(pα) = 1 + α. Therefore if n = p
α11 · · ·pαrr ,
h(n) ={∏
ν;pν≡1 (mod 3)(1 + αν), if all αν for which pν ≡ 1 (mod 3) are even,0, otherwise.
Comments 33 It can be proved that 6h(n) is the number of integer solutions of theequation
x2 − xy + y2 = n.
In the same way, we can prove that the difference between the number τ4k+1(n) ofdivisors of the form 4k + 1 and the number τ4k−1(n) of divisors of the form 4k − 1is also positive. Moreover, 4(τ4k+1(n) − τ4k−1(n)) is the number of solutions of theDiophantine equation
x2 + y2 = n.
Comments 34 The positivity result holds more generally when one compares thenumbers of divisors of the form 4k ± 1, 6k ± 1, 8k ± 1, 12k ± 1, 24k ± 1. However,the analogous result do not hold for any other congruences.
Comments 35 The number of solutions in integers of quadratic equations appears inmany arithmetic questions. The oldest and perhaps one of the most famous unsolvedproblems is the congruent numbers problem. Specifically, one asks to determine allintegers that are the areas of right triangles with rational sides, which are calledcongruent numbers. For instance, n = 6 is the area of the right triangle of sides 3, 4,and 5. It is less obvious that 157 is a congruent number, where the simplest solutionfor the sides of the associated right triangle reads
6803298487826435051217540
411340519227716149383203,
411340519227716149383203
21666555693714761309610,
224403517704336969924557513090674863160948472041
8912332268928859588025535178967163570016480830.
It is known that congruent numbers are those d for which the equation
dy2 = x3 − x
has infinitely many rational solutions.
Deep results of Tunnell imply the following conjecture, which would be a completecharacterization of congruent numbers. Let d be square-free. Denote by N(d) (andrespectively M(d)) the number of integer solutions of
2x2 + y2 + 8z2 = d, and respectively 2x2 + y2 + 32z2 = d, for odd d,
5 Number Theory Solutions 81
4x2 + y2 + 8z2 = d
2, and respectively 4x2 + y2 + 32z2 = d
2, for even d.
Then d is a congruent number if and only if
N(d) = 2M(d).
This is a particular case of the Birch–Swinnerton-Dyer conjecture, considered oneof the most important unsolved problems in arithmetic, and included as one of theseven Millennium Problems (for which the Clay Mathematics Institute will award onemillion dollars for a solution). The interested reader might consult
• N. Koblitz: Introduction to Elliptic Curves and Modular Forms, second edition,Graduate Texts in Mathematics, 97, Springer-Verlag, New York, 1993.
Problem 1.69. A number N is called deficient if σ(N) < 2N and abundant ifσ(N) > 2N .
1. Let k be fixed. Are there any sequences of k consecutive abundant numbers?2. Show that there are infinitely many 5-tuples of consecutive deficient numbers.
Solution 1.69. 1. Let pn denote the nth prime number. It is known that the infiniteproduct
∞∏
n=1
(1 + 1
pn
)
is infinite. Thus, for any m, there exists an integer N(m) such that
N(m)∏
n=m
(1 + 1
pn
)> 2.
In particular, this shows that the number t = ∏N(m)n=m pn is abundant, because σ(t) =
t∏N(m)n=m
(1 + 1
pn
). Consider now the sequence defined by
m1 = 1, mj+1 = N(mj )+ 1, for j ≥ 1, and tj =N(mj )∏
n=mjpn.
Then the numbers tj are pairwise relatively prime, and therefore for every k thereexists some integer L (depending on k) with the property that L + j ≡ 0 (mod tj ),for all j ∈ {1, . . . , k}. Observe that any multiple of an abundant number is abundantitself. This shows that each of the numbers L+ 1, . . . , L+ k is abundant.
2. It is well known that
limn→∞
(1 + 2
n log n
)n= 1
and limn→∞ pnn log n = 1. Take n large enough so that
82 5 Number Theory Solutions
(1 + 2
n log n
)n< 1.01, and pn >
1
2n log n > 60.
Let us next consider
Qn =n−1∏
i=1
pj , and Mn = 60pn + 1.
We set xi = Mn+i−1i
for i ∈ {1, 2, 3, 4, 5}. Because n > 3, it is easy to see that everyxi is relatively prime to Qn. Therefore each xi can be factored into prime factors
xi =ti∏
j=1
pnji ,
where each nji ≥ n (because otherwise, it would have common factors withQn) andthe number of factors ti is less than or equal to n (otherwise, xi would be greater thanMn).
Since 1 + 1pnji
≤ 1 + 1pn
and for prime p we have σ(ps)/ps ≤ (σ (p)/p)s , we
conclude that
σ(xi)
xi≤
(1 + 1
pn
)n<
(1 + 2
n log n
)n< 1.01
and therefore
σ(Mn + i − 1)
Mn+i= σ(i)
i· σ(xi)xi
< 1.75 · 1.01 < 2 .
Thus Mn,Mn + 1,Mn + 2,Mn + 3,Mn + 4 are deficient numbers.
Problem 1.70. Does there exist a nonconstant polynomial an2 + bn+ c with integercoefficients such that for any natural numberm, all its prime factorspi are congruent toto 3 modulo 4? Prove that, for any nonconstant polynomial f with integer coefficientsand any m ∈ Z there exist a prime number p and a natural number n such that pdivides f (n) and p ≡ 1 (mod m).
Solution 1.70. 1. We write 4af (n) = (2an+b)2+4ac−b2. If a = 0 or 4ac−b2 = 0,it is obvious that there are infinitely many prime numbers 4m+1 that divide f (n). Letus then assume that a(4ac − b2) �= 0. Set p1, . . . , pm for the odd prime divisors ofa(4ac − b2). The Chinese remainder theorem and Dirichlet’s theorem on arithmeticprogressions imply the existence of a prime p satisfying p ≡ 1 (mod pi), for
all i and p ≡ 1 (mod 8). The Legendre symbol satisfies the identities(pip
)=
(ppi
)=
(2p
)=
(−1p
)= 1, which imply that there exists an integer m such that
m2 ≡ b2−4ac (modp). Since gcd(2a, p) = 1, the congruence 2an+b ≡ m(modp)has some solution n and then p divides f (n). The answer is therefore negative.
5 Number Theory Solutions 83
2. Assume that f is irreducible. Let α1, . . . , αk be the roots of f and let ω be aprimitive mth root of unity. The field K = Q(ω, α1, . . . , αk) is a normal extensionof Q. The Cebotarev density theorem says that the density of rational primes thatcompletely split in K is 1/[K : Q], and thus nonzero. Let p be such a prime thatmoreover, does not divide the discriminant of K . Then p splits completely in everysubfield of K , in particular in Q(α1). From Dirichlet’s unit theorem, it follows thatthe congruence f (x) ≡ 0 (mod p) has k distinct solutions, since p does not dividethe discriminant of Q(α1). Let �m be the cyclotomic polynomial of degree m. Asimilar argument concerning Q(ω) shows that the congruence �m(x) ≡ 0 (mod p)has exactly φ(m) roots. Let r be one of them. Then the order of r in Z/pZ is exactlym and thus m divides p − 1. Therefore p has the required properties.
Comments 36 If we required the prime factors to be congruent to 1 modulo 4, thenthe answer would be positive. In fact, for any integer n, the prime divisors of 4n2 + 1are of the form 4m+ 1.
6
Algebra and Combinatorics Solutions
6.1 Algebra
Problem 2.1. Set Sk,p = ∑p−1i=1 i
k , for natural numbers p and k. If p ≥ 3 is primeand 1 < k ≤ p − 2, show that
Sk,p ≡ 0 (mod p).
Solution 2.1. Consider the symmetric polynomial P(x) = xk1 + · · · + xkp−1. Ac-cording to the fundamental theorem of symmetric polynomials, there exists G ∈Z[x1, . . . , xp−1] such that P(x) = G(σ1, . . . , σp−1), where σi(x1, . . . , xp−1) arethe fundamental symmetric polynomials
σi(x1, . . . , xp−1) =∑
1≤j1<···<ji≤p−1
xj1 · · · xji .
Furthermore, the largest n with the property that σn actually appears in the develop-ment ofG satisfies n ≤ degP . In particular, if k ≤ p−2, thenP = G(σ1, . . . , σp−1).
Let us show first that σi(1, 2, . . . , p − 1) ≡ 0 (mod p). In fact, the theorem ofFermat tells us that ap−1 ≡ 1 (mod p), for any number a �≡ 0 (mod p). Thus thepolynomial Q = xp−1 − 1 ∈ Z/pZ [x] with coefficients in Z/pZ vanishes for allnonzero classes {1 . . . , p−1} modulop, and so these are all the roots ofQ. Therefore,Q decomposes in Z/pZ [x] as Q = (x − 1) · · · (x − (p − 1)). By identifying thecoefficients of the right- and left-hand sides, we obtain σi(1, . . . , p−1) ≡ 0 (mod p),as claimed. Finally, Sk,p = P(1, . . . , p−1) = G(0, 0, 0, . . . , 0) ≡ 0 (modp), whichproves the claim.
Observe that if k = p − 1, then we have ip−1 ≡ 1 (mod p) and hence Sp−1,p ≡−1 (mod p). In particular, we have
Sk,p ={
0, if k �≡ −1 (mod p − 1),−1, if k ≡ −1 (mod p − 1).
86 6 Algebra and Combinatorics Solutions
Problem 2.2. LetP = a0 +· · ·+anxn andQ = b0 +· · ·+bmxm be two polynomialswith m ≤ n. Then, deg gcd(P,Q) ≥ 1 if and only if there exist two polynomials Kand L such that deg K ≤ m− 1, deg L ≤ n− 1, andK · P = L ·Q. Prove that thisis equivalent to the vanishing of the following (n+m)× (n+m) determinant:
det
⎛
⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝
a0 a1 a2 · · · am−1 · · · an−1 an 0 0 · · · 00 a0 a1 · · · am−2 · · · an−2 an−1 an 0 · · · 0......
......
......
......
...
0 0 0 · · · a0 · · · an−m−1 an−m an−m+1 · · · an0 0 0 · · · 0 · · · an−m−2 an−m−1 an−m · · · an−1......
......
......
......
0 0 0 · · · 0 0 a0 a1 a2 · · · amb0 b1 · · · bm−1 bm 0 0 0 0 · · · 00 b0 · · · bm−2 bm−1 bm 0 0 0 · · · 0......
......
......
......
...
0 0 · · · b0 b1 · · · · · · · · · 0 · · · 0
⎞
⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠
= 0.
Solution 2.2. Let us write
P = an∏
∑δi=n
(x − xi)δi , Q = bm
∏
∑γi=m
(x − xi )γi ,
L = cm−1
∏
∑ξi=n−1
(x − yi)ξi , K = dm−1
∏
∑ψi=m−1
(x − yi )ψi ,
and then consider the identity P ·K = Q ·L, each side being decomposed into linearterms.According to the pigeonhole principle, after the reduction of the common terms,on each side, left and right, there remains some polynomial of degree ≥ n+m− (n+m− 1) = 1.
Conversely, if P = P0 · P1,Q = P0 ·Q1, then Q1 · P = P1 ·Q.For the second assertion, set K = ∑
uixi and L = ∑
wixi . By identifying the
coefficients of P ·K and Q · L, one obtains the identities∑
j+k=iajuk =
∑
j+k=ibjwk.
Alternatively,
D · (u0, u1, . . . , un−1,−w0,−w1, . . . ,−wm−1)� = 0,
whereD denotes the matrix from the claim. SinceK and L are nonzero, this impliesthat detD = 0.
Problem 2.3. Prove that if P,Q ∈ R[x, y] are relatively prime polynomials, thenthe system of equations
P(x, y) = 0,Q(x, y) = 0,
has only finitely many real solutions.
6.1 Algebra 87
Solution 2.3. Assume that the value of y is a fixed y0, so that there exists at leastone solution (x, y0). Then the polynomials P(x, y0) and Q(x, y0) from R[x] haveat least one common root, and hence a nontrivial common divisor. According to thepreceding problem, this amounts to the vanishing of the determinant of some matrixDy0 = D(P (x, y0),Q(x, y0)) obtained from the coefficients.
Further, the determinant of Dy0 is a nonzero polynomial in y0. In fact, if Dywere identically zero, then P andQwould have a common divisor, contradicting ourassumptions. Therefore the number of such y0 for which a solution (x, y0) exists isfinite.
In a similar way, the number of those x0 for which there exist solutions (x0, y) isfinite, by the same argument. This implies that the number of solutions is finite.
Problem 2.4. Let a, b, c, d ∈ R[x] be polynomials with real coefficients. Set
p =∫ x
1ac dt, q =
∫ x
1ad dt, r =
∫ x
1bc dt, s =
∫ x
1bd dt.
Prove that ps − qr is divisible by (x − 1)4.
Solution 2.4. Let p, q, r, s be smooth functions such that p(u) = q(u) = r(u) =s(u) = 0 and det
(p′ q ′r ′ s′
)(u) = 0. Then the function det
(p qr s
), together with its first
three derivatives, vanishes at u. The proof is immediate. The given p, q, r, s satisfythe previous conditions at u = 1, and hence the claim.
Problem 2.5. 1. Find the minimum number of elements that must be deleted fromthe set {1, . . . , 2005} such that the set of the remaining elements does not contain twoelements together with their product.
2. Does there exist, for any k, an arithmetic progression with k terms in the infinitesequence
1,1
2, . . . ,
1
2005, . . . ,
1
n, . . .?
Solution 2.5. 1. If we extract the numbers 1, 2, 3, . . . , 44 and then choose x, y ∈{45, . . . , 2005}, then xy ≥ 452 > 2005, and thus it cannot belong to the given set.Therefore the number of elements to be deleted is at most 43.
Consider now the triples (2, 87, 2 · 87), (3, 86, 3 · 86), . . . , (44, 45, 44 · 45) con-sisting of elements of the given set. We must delete at least one number from eachtriple in order that the remaining set be free of products. Thus the minimum numberof elements to be deleted is 43.
2. Yes, it does. Take for instance 1k! ,
2k! , . . . ,
kk! .
Problem 2.6. Consider a set S of n elements and n+ 1 subsets M1, . . . ,Mn+1 ⊂ S.Show that there exist r, s ≥ 1 and disjoint sets of indices {i1, . . . , ir}∩{j1, . . . , js} = ∅such that
r⋃
k=1
Mik =s⋃
k=1
Mjk .
88 6 Algebra and Combinatorics Solutions
Solution 2.6. Assume that S = {a1, a2, . . . , an}. We associate to the subset Mj thevector vj ∈ R
n having the components vj = (v1j , v2j , . . . , vnj ), given by
vij ={
1, if ai ∈ Mj,
0, if ai �∈ Mj.
Since we have n + 1 vectors v1, . . . , vn+1 in Rn, there exists a nontrivial linear
combination of them that vanishes. We separate the positive coefficients and thenegative coefficients and write down this combination in the form
∑
i∈Iλivi =
∑
j∈Jλj vj ,
where I, J are disjoint sets of indices and λk > 0, for k ∈ I ∪ J . We claim now that
∪i∈IMi = ∪j∈JMj .
In fact, assume that ak ∈ ∪i∈IMi and thus ak ∈ Mi for some i ∈ I . Then the kthcoordinate of the vector vi is nonzero. All coordinates of the vectors vs are non-negative and the coefficients λi are positive, which implies that the kth coordinate ofthe vector
∑i∈I λivi is nonzero. Since the kth coordinate of
∑j∈J λj vj is nonzero,
there should exist some j ∈ J for which the kth coordinate of vj is nonzero andhence ai ∈ Mj ⊂ ∪j∈JMj . This proves that ∪i∈IMi ⊂ ∪j∈JMj , and by symmetry,we have equality.
Problem 2.7. Let p be a prime number, and A = {a1, . . . , ap−1} ⊂ Z∗+ a set of
integers that are not divisible by p. Define the map f : P(A) → {0, 1, . . . , p − 1}by
f ({ai1 , . . . , aik }) =k∑
p−1
aip (mod p), and f (∅) = 0.
Prove that f is surjective.
Solution 2.7. Let C0 = ∅, Cn = {a1, . . . , an}, for n ≤ p − 2. Set Pn = {f (β)|β ⊂Cn}, and bn = f ({an}) �≡ 0. One obtains easily the following inductive descriptionof the sets Pn:
Pn+1 = {r + bn+1|r ∈ Pn ∪ {0}} ∪ Pn.Furthermore, if Pn = Pn+1, then bn+1 + r ∈ Pn for all r ∈ Pn. In particular,bn+1 ∈ Pn, and by induction, kbn+1 ∈ Pn, for any k. Since p is prime and bn+1 �= 0,we obtain Pn = {0, 1, . . . , p − 1}.
Assume now that Pn �= Pn+1 for all n. Since card P0 = 1 and Pn ⊂ Pn+1, wederive that card Pn ≥ n+ 1, and hence Pp−1 = P .
Problem 2.8. Consider the function Fr = xr sin rA + yr sin rB + zr sin rC, wherex, y, z ∈ R, A + B + C = kπ , and r ∈ Z+. Prove that, if F1(x0, y0, z0) =F2(x0, y0, z0) = 0, then Fr(x0, y0, z0) = 0, for all r ∈ Z+.
6.1 Algebra 89
Solution 2.8. Consider the complex numbers
u = x0(cosA+ i sinA), v = y0(cosB + i sinB), w = z0(cosC + i sinC).
We denote the argument of the complex number z by arg z and its imaginary partby �z. We have arg u + arg v + arg w = kπ, � u + � v + � w = 0, and� u2 + � v2 + � w2 = 0. Now � z = 0 means that z is real, so that � z2 = 0.Thus 2�(uv + vw + wu) = �((u + v + w)2) − �(u2 + v2 + w2) = 0. Next,arg uvw = arg u+ arg v + arg w = kπ , and so � uvw = 0.
Let us consider the polynomial Pn(x1, x2, x3) = xn1 + xn2 + xn3 . According to thefundamental theorem of symmetric polynomials, we can write Pn as a polynomial inthe fundamental symmetric polynomials:
Pn(x1, x2, x3) = G(σ1, σ2, σ3),
whereG ∈ R[σ1, σ2, σ3] is a real polynomial andσi , i ∈ {1, 2, 3}, are the fundamentalsymmetric polynomials in three variables, namely
σ1(x1, x2, x3) = x1 + x2 + x3, σ2(x1, x2, x3) = x1x2 + x2x3 + x3x1,
σ3(x1, x2, x3) = x1x2x3.
Notice now that σ1(u, v,w) = u + v + w ∈ R, σ2(u, v,w) = uv + vw + wu ∈R, σ3(u, v,w) = uvw ∈ R, which implies that Pn(u, v,w) ∈ R, and thereforeFr(x0, y0, z0) = �(Pn(u, v,w)) = 0.
Problem 2.9. Let T (z) ∈ Z[z] be a nonzero polynomial with the property that|T (ui)| ≤ 1 for all values ui that are roots of P(z) = zn − 1. Prove that eitherT (z) is divisible by P(z), or else there exists some k ∈ Z+, k ≤ n − 1, such thatT (z) ± zk is divisible by P(z). The same result holds when instead of P(z), weconsider zn + 1.
Solution 2.9. Let us write T (z) = A(z)P (z) + B(z), where deg B ≤ n − 1 andA,B ∈ Z[x]. Assume that T (z) is not divisible by P(z), and so B �= 0. Set k for themultiplicity of the root 0 in B, meaning that B(z) = zkC(z), where C(0) �= 0. Wehave then |C(ui)| = |B(ui)| = |T (ui)| ≤ 1.
If we set Qd(z) = zd , then we can compute easily the sum of roots of unity as
d∑
i=1
Qd(ui) = 0, if 1 ≤ d ≤ n− 1.
We want to compute this kind of sum for the polynomial C. Since degC ≤ n − 1,the previous identities show that only the degree-zero part of C has a nontrivialcontribution, and thus
n∑
i=1
C(ui) = nC(0).
This implies that
90 6 Algebra and Combinatorics Solutions
0 < |C(0)| ≤ 1
n
∣∣∣∣∣
n∑
i=1
C(ui)
∣∣∣∣∣≤ 1
n
n∑
i=1
|C(ui)| ≤ 1.
Now C(0) ∈ Z and thus C(0) = ±1, which implies that either C(ui) = 1, for all i,or else C(ui) = −1, for all i. Since deg C ≤ n− 1, we derive that either C(z) = 1or else C(z) = −1. In particular, T (z)± zk is divisible by P(z).
The case in which P(z) is replaced by zn + 1 is similar.
Problem 2.10. 1. If the map x �→ x3 from a group G to itself is an injective grouphomomorphism, then G is an abelian.
2. If the map x �→ x3 from a groupG to itself is a surjective group homomorphism,then G is an abelian.
3. Find an abelian group with the property that x �→ x4 is an automorphism.4. What can be said for exponents greater than 4?
Solution 2.10. 1. We have a3b3 = (ab)3, and thus a2b2 = (ba)2, whence a4b4 =(a2)2(b2)2 = (b2a2)2 = (ab)4. Using again that a3b3 = (ba)3, we get (ab)3 =(ba)3. Since the map is injective, we have ab = ba.
2. We continue from above, a3b3 = (ba)3 = (ba)2ba = a2b2(ba); thus ab3 =b3a. Since the homomorphism is surjective, b3 can take any value in G; hence therelation above shows that G is abelian.
3. Take any nonabelian group of order 8, for instance the dihedral group D4.4. For m > 3, there exists a nonabelian group Gm such that x �→ xm is the
identity. For instance, if p is not of the form 2n + 1, then let p be an odd prime factorof m− 1 and Gm the group of matrices of the form
M =⎛
⎝1 a b0 1 c0 0 1
⎞
⎠
with arbitrary a, b, c ∈ Z/pZ. Then Gm is a nonabelian group of order p3 such thatthe order of every element �= e is p.
Problem 2.11. Let V be a vector space of dimension n > 0 over a field of character-istic p �= 0 and let A be an affine map A : V → V . Prove that there exist u ∈ V and1 ≤ k ≤ np such that Aku = u.
Solution 2.11. We write Ax = Bx + a, and thus Akx = Bkx + Cka, where Ck =Bk−1 + · · · +B + I . We use (B − I )Ck = Bk − I to rewrite the equation to solve as(B − I )Cku = Cka. There exists a solution if and only if a ∈ ker Ck + Im(B − I ).We suppose B− I is singular. The restriction of B to the eigenspace S correspondingto the eigenvalue 1 can be written BS = I +N , where N is nilpotent. Therefore
Ck|S =k−1∑
i=0
(I +N)i =k−1∑
i=0
i∑
j=0
Cji N
j =k−1∑
j=0
(k−1∑
i=jCji
)Cji N
j =k−1∑
j=0
Cj+1k Nj .
6.1 Algebra 91
If p divides k, then Ck|S = Nk−1, and thus if k − 1 ≥ dim S, then Ck|S = 0.Therefore, S ⊂ ker Ck if k = lp ≥ 1 + dim S.
Since Im(B − I ) ⊃ Im(B − I )n, which is complementary to S, we can takek = (n − 1)p if (n − 1)p ≥ dim S + 1, in which case Im(B − I ) + ker Ck = V .Otherwise, we have either n = 1, or n = 2, p = 2, and dim S = 2. If n = 1, then takek = p. If n = 2 = p, then take k = 2 if dim S = 1 or B = I , and k = 4 otherwise.In this situation, k = np is convenient, while in the former case k = (n−1)p suffices.This proves the claim.
Problem 2.12. Find the cubic equation the zeros of which are the cubes of the rootsof the equation x3 + ax2 + bx + c = 0.
Solution 2.12. It is known that if the matrix M has eigenvalues λ1, . . . , λn and if Fis a polynomial, then F(M) has eigenvalues F(λ1), . . . , F (λn). We consider now thematrix
M =⎛
⎝0 0 −c1 0 −b0 1 −a
⎞
⎠ ,
which has the characteristic polynomial x3 + ax2 + bx + c. The characteristic poly-nomial of M3 will then have as roots the cubes of the given polynomial. We nowcompute easily the characteristic polynomial ofM3 as x3 + (a3 −3ab+c)x2 + (b3 −3ab + 3c2)x + c3 = 0.
Problem 2.13. Assume that the polynomials P,Q ∈ C[x] have the same roots, pos-sibly with different multiplicities. Suppose, moreover, that the same holds true for thepair P + 1 and Q+ 1. Prove that P = Q.
Solution 2.13. LetA andB be the sets of roots of P and P +1, respectively. Supposethat deg P = m ≥ n = deg Q. Obviously, A and B should be disjoint. It followsthat P ′ has at leastm− r roots from the set A and at leastm− s roots from the set B,because multiple roots of P are roots of P ′. Since deg P ′ = m − 1, we derive that(m − r) + (m − s) ≤ m − 1 and thus r + s > m. Now, the union A ∪ B has r + s
elements, because these sets are disjoint. On the other hand, each element of A ∪ Bis a root of the polynomial P −Q, of degreem. Thus P −Q has at leastm+ 1 roots,and thus it vanishes identically.
Problem 2.14. Determine r ∈ Q, for which 1, cos 2πr, sin 2πr are linearly depen-dent over Q.
Solution 2.14. We consider 0 ≤ r < 1. We suppose that there exists a linear relationa + b cos 2πr + c sin 2πr = 0, with a, b, c, r ∈ Q. This implies that
(b2 + c2) cos2 2πr + 2ab cosπr + a2 − c2 = 0,
i.e., cos 2πr satisfies a quadratic equation with rational coefficients. The same holdsfor sin 2πr .
92 6 Algebra and Combinatorics Solutions
Consider now ξ = cos 2πr + i sin 2πr . If n is the smaller denominator of r , thenξ is a primitive root of nth order.
On the other hand, since both cos 2πr and sin 2πr satisfy quadratic equations,then ξ satisfies an equation of order 4, with rational coefficients.
We know that ξn = 1. Further, the irreducible polynomial of smallest degree thatdivides xn − 1 is 1 + x + · · · + xϕ(n), of degree ϕ(n), where ϕ(n) is Euler’s totientfunction.
Our previous discussion implies that ϕ(n) must be a divisor of 4, and henceϕ(n) ∈ {1, 2, 4}. Therefore, we have the following cases:
1. ϕ(n) = 1 and hence n ∈ {1, 2}, and so r ∈ {0, 1
2
}.
2. ϕ(n) = 2, and thus n ∈ {3, 4, 6}, whence r ∈{
14 ,
34 ,
13 ,
23 ,
16 ,
56
}.
3.ϕ(n) = 4, and thus n ∈ {5, 8, 10, 12}, whence r ∈{
18 ,
38 ,
58 ,
78 ,
112 ,
512 ,
712 ,
1112
}.
We omitted in the list above those r having denominator 5 or 10 because
sin 2π5 =
√12 (5 + √
5) does not satisfy any quadratic equation with rational coef-
ficients. Therefore, r ∈ {x|x = m
24 , gcd(m, 24) �= 1,m ∈ Z}.
Problem 2.15. 1. Prove that there exista, b, c ∈ Z, not all zero, such that |a|, |b|, |c| <106
∣∣∣a + b√
2 + c√
3∣∣∣ < 10−11.
2. Prove that if 0 ≤ |a|, |b|, |c| < 106, a, b, c ∈ Z, and at least one of them is
nonzero, then∣∣∣a + b
√2 + c
√3∣∣∣ > 10−21.
Solution 2.15. 1. Consider the set S ={r + s
√2 + t
√3}
r,s,t∈{0,1,...,106−1}. Then
card S = 1018. Set d = (1 + √
2 + √3)106. Then all elements x ∈ S are bounded
by d , i.e., 0 ≤ x < d. One divides the interval [0, d] into 1018 − 1 equal intervals[(k − 1)e, ke], where e = 10−18d and k = 1, . . . , 1018 − 1. By the pigeonholeprinciple, there exist two elements x, y ∈ S lying in the same interval, and hencesatisfying
|x − y| ≤ e < 10−11
2. Let F1 = a + b√
2 + c√
3, F2 = a + b√
2 − c√
3, F3 = a − b√
2 + c√
3,F4 = −a + b
√2 + c
√3. Now, 1,
√2,
√3 are linearly independent over Z, and thus
Fi �= 0, for all 1 ≤ i ≤ 4. Further, F = F1F2F3F4 ∈ Z, since the product ofconjugates contains no more square roots. This implies that |F1| ≥ 1/|F2F3F4| >10−21, because 1/|Fi | > 10−7.
Problem 2.16. Prove that if n > 2, then we do not have any nontrivial solutions forthe equation
xn + yn = zn,
where x, y, z are rational functions. Solutions of the form x = af, y = bf, z = cf ,wheref is a rational function anda, b, c are complex numbers satisfyingan+bn = cn,are called trivial.
6.1 Algebra 93
Solution 2.16. Any rational solution yields a polynomial solution, by clearing thedenominators. Let further (f, g, h) be a polynomial solution for which the degreer = max{deg(f ), deg(g), deg(h)} is minimal among all polynomial solutions, andwhere r > 0.
Consider the root of unity ξ = exp( 2πin
)and assume that f, g, and h are relatively
prime. The equation can be written as
n−1∏
j=0
(f − ξjh) = gn.
Now, gcd(f, h) and gcd(f, g) are relatively prime, so that f − ξjh and f − ξkh arerelatively prime when k �= j . Since the factorization of polynomials is unique, wemust have g = g1 · · · gn−1, where gj are polynomials satisfying gnj = f − ξjh.
Consider now the set {f − h, f − ξh, f − ξ2h}. Since n > 2, these elementsbelong to the two-dimensional space generated by f and h over C. Thus there existsa vanishing linear combination with complex coefficients in these three elements.Thus, there exist ai ∈ C so that a0g
n0 + a1g
n1 = a2g
n2 . We then set hj = n
√ajgj , and
observe thathn0 + hn1 = hn2 .
Moreover, the polynomials hi and hj are relatively prime if i �= j and max deg(hi) <r , which contradicts our choice of r . This proves the claim.
Problem 2.17. A table is an n × k rectangular grid drawn on the torus, every boxbeing assigned an element from Z/2Z. We define a transformation acting on tablesas follows. We replace all elements of the grid simultaneously, each element beingchanged into the sum of the numbers previously assigned to its neighboring boxes.Prove that iterating this transformation sufficiently many times, we always obtain thetrivial table filled with zeros, no matter what the initial table was, if and only if n = 2p
and k = 2q , for some integers p, q. In this case, we say that the respective n× k gridis nilpotent.
Solution 2.17. Consider the m×m square matrices (m ≥ 2)
Dm =
⎛
⎜⎜⎜⎜⎜⎜⎜⎝
0 1 0 0 · · · 0 11 0 1 0 · · · 0 00 1 0 1 · · · 0 0.....................
0 0 0 0 · · · 0 11 0 0 0 · · · 1 0
⎞
⎟⎟⎟⎟⎟⎟⎟⎠
.
We will write ≡ below if the equality holds (only) for matrices of entries modulo2. Then the transformation T from the statement acts on the vector spaceMn,k(Z/2Z)
of n× k matrices with entries in Z/2Z as follows:
T (X) = DnX +XDk, for X ∈ Mn,k(Z/2Z).
94 6 Algebra and Combinatorics Solutions
By induction on s, we have that
T 2s (X) ≡ D2sn X +XD2s
k .
Thus there exists N such that T N(X) ≡ 0 if and only if there exists some s such thatD2sn X ≡ XD2s
k .Moreover, we can prove easily that for two matricesA andB of appropriate sizes,
we have AX ≡ XB for all X ∈ Mn,k(Z/2Z) if and only if A and B are multiples ofthe identity by the same scalar. This follows, for instance, if we take X having all butone column (or line) trivial. This assertion is a special case of Schur’s lemma.
Thus T is nilpotent if and only if there exist s and β ∈ Z/2Z such thatD2sn = β ·1n
and D2sk = β · 1k . Thus the n× k grid is nilpotent if and only if the n× 1 and 1 × k
grids are nilpotent.Let us analyze the case of the n× 1 grid. Set a = (a1, . . . , an) and a[k] = Dkna =(
a[k]1 , . . . , a
[k]n
). By induction, we have
a[2k]s ≡ as+2k−1 + as−2k−1 ,
where the indices are modulo n. In particular, if n = 2q , then D2qn = 0 and the grid
is nilpotent. This proves the “if” part of the statement.Assume next that n is odd and there exists some r such that a[r] ≡ 0, for any a.
Take such an r that is minimal. If n ≥ 3, then r ≥ 3. We claim now that a[r−1]s ≡ 1 for
all s. This is true at least for one s, since r is minimal. Since a[r]s ≡ a
[r−1]s+1 + a
[r−1]s−1 ,
we find that the terms corresponding to even indices (modulo n) are all equal. But nis odd and thus all terms are equal. Further, we have a[r−1]
s = a[r−2]s−1 + a
[r−2]s+1 . We
obtain then that
n ≡n∑
s=1
a[r−1]s =
n∑
s=1
(a
[r−2]s−1 + a
[r−2]s+1
) = 2n∑
s=1
a[r−2]s ≡ 0 (mod 2),
which is a contradiction. This proves actually that a[r] ≡ 0 when n is odd if and onlyif a ∈ {(0, 0, . . . , 0), (1, 1, . . . , 1)}.
Finally, consider n = 2mh, with h odd. For a vector a of length n, we set
�s(a) = (as, as+2m, . . . , as+(h−1)2m).
We observe now that�s(T
2m(a)) = T (�s(a))
because a[2m+1]s = as + as+2m . Since the vector �s(a) is of length h (which
is odd), the previous claim shows us that T N(�s(a)) ≡ 0 for some N only if�s(a) ∈ {(0, 0, . . . , 0), (1, 1, . . . , 1)}. Moreover, this happens for all values of s.This condition is trivial only when h = 1. This proves the “only if” part.
Comments 37 This result was stated in the 1980s as an open question in the journalKvant. The same method can be used to show that the n × k grids in the plane are
6.2 Algebraic Combinatorics 95
nilpotent if and only if n = 2p − 1 and k = 2q − 1 for some natural numbers p, q. Anice corollary is the following. Define the sequence (αk) by the recurrence relations
α1 = 0, αk+1 = min(2αk + 2, 2N − 2αk − 1).
Then there exists some k such that αk = N − 1 if and only if the parameterN has theform 2q − 1 for some natural number q. The solution given here is from:
• V. Boju, L. Funar: Iterative processes for Zn2 , Analele Univ. Craiova 15 (1987),
33–38.
6.2 Algebraic Combinatorics
Problem 2.18. Let us consider a four-digit number N whose digits are not all equal.We first arrange its digits in increasing order, then in decreasing order, and finally,we subtract the two obtained numbers. Let T (N) denote the positive difference thusobtained. Show that after finitely many iterations of the transformation T , we obtain6174.
Solution 2.18. Let N have the digits a, b, c, d.1. Assume that a ≥ b = c ≥ d. Then in T (N), the sum of the first and the fourth
digits is 9, as well as the sum of the second and the third. These give five combinationsof four digits that give 6174, by applying T once more.
2. Otherwise, a ≥ b > c ≥ d. Then in T (N), the sum of the digits placed atextremities is 10, while the sum of the middle ones is 8. This gives 25 combinationsall leading to the desired result, namely 6174.
Comments 38 Let a be a positive integer having r digits, not all equal, in base g. Leta′ be the g-adic integer formed by arranging the digits of a in descending order andlet a′′ be that formed by ascending order. Define T (a) = a′ − a′′. Then a is said to beself-producing if T (a) = a. There exists an algorithm for obtaining self-producingnumbers in any base.
Such a fixed point k of T is called a (g-adic) Kaprekar constant if it has the furtherproperty that every r-digit integer a eventually yields k on repeated iteration of Tand moreover, k is fixed by T . D.R. Kaprekar found that 6174 has this nice propertyin a short note published in 1949. Ludington proved that for large r ≥ ng there doesnot exist a Kaprekar constant. Here ng is given by
ng =⎧⎨
⎩
g − 1 + (k − 1)δ, if g = 2k,2k + 1 + (g − k − 2)δ, if g = 2t k + 2t−1 + 1,3k − 4 + (g − 1)δ, if g = 2t + 1,
and δ ∈ {0, 1} is equal to g (mod 2). Further improvements by Lapenta, Ludington,and Prichett showed that there is no Kaprekar constant in base 10 when the numberof digits is r > 4.
96 6 Algebra and Combinatorics Solutions
Hasse and Prichett found that there exists a 4-digit Kaprekar constant in base g ifand only if g = 2n · 5, where n is either 0 or an odd number.
Moreover, the situation is completely understood for 3-digit numbers. If the base gis odd, then there is a Kaprekar constant given by
(r−2
2 , r − 1, r2), which is reached
within r+22 iterations of T (respectively 1, if r = 2). For instance, 495 is a Kaprekar
constant if g = 10. If the base g is even, then there is no Kaprekar constant. Moreprecisely, iterations of T will eventually reach the loop of length 2 formed by the pairof numbers
(r−3
2 , r − 1, r+12
),(r−1
2 , r − 1, r−12
), after at most r+1
2 steps (respectively1 if r = 3).
• K.E. Eldridge and S. Sagong: The determination of Kaprekar convergence andloop convergence of all three-digit numbers, Amer. Math. Monthly 95 (1988),105–112.
• J.F. Lapenta, A.L. Ludington, and G.D. Prichett: An algorithm to determine self-producing r-digit g-adic integers, J. Reine Angew. Math. 310 (1979), 100–110.
• G.D. Prichett, A.L. Ludington, and J.F. Lapenta: The determination of all decadicKaprekar constants, Fibonacci Quart. 19 (1981), 45–52.
• A.L. Ludington: A bound on Kaprekar constants, J. Reine Angew. Math. 310(1979), 196–203.
• H. Hasse and G.D. Prichett: The determination of all four-digit Kaprekar con-stants, J. Reine Angew. Math. 299/300 (1978), 113–124.
Problem 2.19. Find an example of a sequence of natural numbers 1 ≤ a1 < a2 <
· · · < an < an+1 < · · · with the property that everym ∈ Z+ can be uniquely writtenas m = ai − aj , for i, j ∈ Z+.
Solution 2.19. We consider the sequence
a1 = 1, a2 = 2,
a2n+1 = 2a2n,
a2n+2 = a2n+1 + rn,
where rn is the smallest natural number that cannot be written in the form ai − aj ,with i, j ≤ 2n+ 1.
Comments 39 One does not know the minimal growth of such a sequence ak .
Problem 2.20. Consider the set of 2n integers {±a1,±a2, . . . ,±an} and m < 2n.Show that we can choose a subset S such that
1. The two numbers ±ai are not both in S;2. The sum of all elements of S is divisible by m.
Solution 2.20. Let S1, . . . , S2n−1 be the 2n − 1 nonempty distinct subsets of the set{a1, . . . , an}, where ai ≥ 0. Let F(Si) denote the sum of the elements of Si . Then, bythe pigeonhole principle, there exist i, j such that F(Si) ≡ F(Sj ) (modm). Considernext the set S = {Si \ Sj } ∪ −{Sj \ Si}. We derive that F(S) ≡ 0 (mod m).
6.2 Algebraic Combinatorics 97
Problem 2.21. Show that for every natural number n there exist prime numbers pand q such that n divides their difference.
Solution 2.21. Consider the following arithmetic progression: 1, 1+n, 1+2n, . . . , 1+rn, . . . .According to Dirichlet’s theorem, there exist infinitely many prime numbersamong the terms of this progression. Let p, q be two of these prime numbers; thenp = 1 + nr and q = 1 + ns, where r �= s. This yields n(r − s) = p− q, as claimed.
An alternative solution is to consider the set of n arithmetic progressions of ration starting respectively at 0, 1, 2, . . . , n− 1. Since the set of prime numbers is infinite(this is elementary), there exists at least one progression having infinitely many primenumbers among its terms. The argument above settles the claim.
Problem 2.22. An even number, 2n, of knights arrive at King Arthur’s court, eachone of them having at most n − 1 enemies. Prove that Merlin the wizard can assignplaces for them at a round table in such a way that every knight is sitting only next tofriends.
Solution 2.22. 1. We consider the friendship graph G defined below: its verticesare in bijection with the knights and the edges are joining pairs of vertices whoserespective knights are not enemies. The degree of each vertex is at least n. Accordingto Dirac’s theorem, such a graph admits a Hamiltonian cycle, and this yields thewanted assignment of places.
2. Choose an arbitrary assignment of places in which we have two neighbors,A and B, who are enemies. Let us assume that A lies on the right-hand side of B.According to the pigeonhole principle, there exists another pair of enemies, say A, B,who are neighbors, and moreover, A is on the right of B.
Let us switch all the places starting at A (and lying on the right side of A) andending at B, using a symmetry. After such a transformation, the number of enemypairs (A,B) is diminished by 2. Applying such transforms iteratively, one obtains aposition in which all neighbors are friends.
Problem 2.23. Let r, s ∈ Z+. Find the number of 4-tuples of positive integers(a, b, c, d) that satisfy 3r7s = lcm(a, b, c) = lcm(a, b, d) = lcm(a, c, d) =lcm(b, c, d).
Solution 2.23. The numbers a, b, c, d are of the form 3mi7ni , 1 ≤ i ≤ 4, where0 ≤ mi ≤ r and 0 ≤ ni ≤ s. Also, mi = r for at least two values of i, and ni = s forat least two values of i. We have then:
(1) one possibility that mi = r for all four i;(2) 4r possibilities that precisely one mi ∈ {0, . . . , r − 1};(3) C2
4r2 possibilities that exactly two mi ∈ {0, . . . , r − 1}.
Therefore, there are(1 + 4r + 6r2
)possibilities for the mi’s and a similar number
for the ni’s, yielding a total of(1 + 4r + 6r2
)(1 + 4s + 6s2
)possibilities.
Problem 2.24. 1. Let n ∈ Z+ and p be a prime number. Denote by N(n, p) thenumber of binomial coefficients Csn that are not divisible by p. Assume that n iswritten in base p as n = n0 + n1p + · · · + nmp
m, where 0 ≤ nj < p, for allj ∈ {0, 1, . . . , m}. Prove that N(n, p) = (n0 + 1)(n1 + 1) · · · (nm + 1).
98 6 Algebra and Combinatorics Solutions
2. Write k in base p as k = k0 + k1p + · · · + ksps , with 0 ≤ kj ≤ p − 1, for all
j ∈ {0, 1, . . . , s}. Prove that
Ckn ≡ Ck0n0Ck1n1
· · ·Cksns (mod p).
Solution 2.24. 1. It is clear that Ckp ≡ 0 (mod p) for all k ∈ {1, 2, . . . , p}. Thus
(1 + x)p ≡ 1 + xp in (Z/pZ)[x]. By induction, we find that (1 + x)pn ≡ 1 + xp
n
in (Z/pZ)[x] for any natural number n. We have then the following congruences in(Z/pZ)[x]:
(1+x)n = (1+x)n0(1+x)n1p · · · (1+x)nmpm ≡ (1+x)n0(1+xp)n1 · · · (1+xpm)nm.When developing factors on the right-hand side, we obtain a sum of factorsxa0+a1p+a2p
2+···+ampm , with a coefficient that is nonzero modulo p. Since everynumber can be uniquely written in base p, all these factors are distinct and theircoefficients modulo p are nonzero. There are exactly (n0 + 1)(n1 + 1) · · · (nm + 1)such factors, and therefore as many binomials not divisible by p.
2. As observed above, the factor xk appears in the form xa0+a1p+a2p2+···+ampm ;
since k can be uniquely written in base p, we have ai = ki . This implies that the
coefficient of xk is Ck0n0C
k1n1 · · ·Ckjnj , and hence the claim.
Problem 2.25. Define the sequence Tn by T1 = 2, Tn+1 = T 2n − Tn + 1, for n ≥ 1.
Prove that if m �= n, then Tm and Tn are relatively prime, and further, that
∞∑
i=1
1
Ti= 1.
Solution 2.25. We prove by induction that Tn+1 = 1 + T1T2 · · · Tn. In fact, we have
Tn+1 = T 2n − Tn + 1 = Tn(Tn − 1)+ 1 = Tn(Tn−1Tn−2 · · · T1)+ 1 = 1 + T1 · · · Tn.
Then takem < n. Therefore, Tm divides T1 · · · Tn−1 = Tn−1 and thus gcd(Tm, Tn) =1. Further, we prove by induction that
n∑
i=1
1
Ti= 1 − 1
Tn+1 − 1.
In fact,
k+1∑
i=1
1
Ti= 1 − 1
Tk+1 − 1+ 1
Tk+1= 1 − 1
Tk+1(Tk+1 − 1)= 1 − 1
Tk+2 − 1.
Since Tn tends to infinity, we therefore obtain∑∞i=1
1Ti
= 1.
6.2 Algebraic Combinatorics 99
Problem 2.26. Let α, β > 0 and consider the sequences
[α], [2α], . . . , [kα], . . . ; [β], [2β], . . . , [kβ], . . . ,
where the brackets denote the integer part. Prove that these two sequences takentogether enumerate Z+ in an injective manner if and only if
α, β ∈ R \ Q and1
α+ 1
β= 1.
Solution 2.26. Set AN = {1, 2, . . . , N}; take k maximal such that kα < N + 1and l maximal such that lβ < N + 1. Therefore, the following inequalities hold:[Nα
] ≤ k ≤ [N+1α
]and
[Nβ
]≤ l ≤
[N+1β
].
Since AN is injectively enumerated by the two sequences, we have k + l = N ,and hence [
N
α
]+
[N
β
]≤ N ≤
[N + 1
α
]+
[N + 1
β
].
Letting N go to infinity, we obtain 1α
+ 1β
= 1.
If α ∈ Q, then also β ∈ Q. In this case, write α = mn
and β = pq
. It follows that[αnqp] = [βmqn], which contradicts the injectivity assumption. Therefore α �∈ Q
and β �∈ Q.Conversely, we have
N+1 = N + 1
α+N + 1
β>
[N + 1
α
]+[N + 1
β
]>N + 1
α+N + 1
β−2 = N−1,
whence [N + 1
α
]+
[N + 1
β
]= N.
In particular, using the notation from above, we have k + l = N .If the two sequences enumerate AN in a surjective manner, then they will also
enumerateAN injectively, because k+l = N . It suffices then to prove the surjectivity.Let us assume the contrary. Then there exist u, x, y ∈ Z+ such that xα < u <
u + 1 < (x + 1)α and yβ < u < u + 1 < (y + 1)β. Dividing by α, β respectivelyand adding up the inequalities, we obtain
x + y <u
α+ u
β<u+ 1
α+ u+ 1
β< x + 1 + y + 1,
which amounts tox + y < u < u+ 1 < x + 2 + y.
This is false, because x, y, u are integers, and our claim follows.
Problem 2.27. We say that the sets S1, S2, . . . , Sm form a complementary system ifthey make a partition of Z+, i.e., every positive integer belongs to a unique set Si . Letm > 1 and α1, . . . , αm ∈ R+. Then the sets
100 6 Algebra and Combinatorics Solutions
Si = {[nαi], where n ∈ Z+}form a complementary system only if
m = 2, α−11 + α−1
2 = 1, and α1 ∈ R \ Q.
Solution 2.27. If all αj < 1, then the collection Sj contains twice some number. Let1 ∈ S1, and thus [α1] = 1, α1 �= 1. We prove first that S1 = {[nα1]; n ∈ Z+} andT = {[nβ − ε]; n ∈ Z+} form a complementary system, where
β = α1
α1 − 1, and ε =
{(2(a − c))−1, if α1 = a
c∈ Q,
0, otherwise.
Let � = {nα1, nβ − ε; n ∈ Z+}. It is sufficient to show that for any integer M > 1,we have the formula
N = card({x ∈ �; x < M}) = M − 1.
Now, nα1 < M is equivalent to nα1 + δ < M , where δ = (2c)−1 if α1 = a/c ∈ Q,
while δ = 0, if α ∈ R \Q. The maximum value for n is therefore[M−δα1
]. In a similar
way, the number of elements of the form nβ − ε that are less than M is[M+εβ
]. This
implies that N = [(M − δ)/α1] + [(M + ε)/β]. Now we have
M − δ
α1− 1 <
[M − δ
α1
]<M − δ
α1,
M + ε
β− 1 <
[M + ε
β
]<M + ε
β,
and by summing up the two inequalities, we obtain M − 2 < N < M , and thereforeN = M − 1, as claimed.
Now let k be the smallest integer k �∈ S1. Then there exists some α2 such thatk = [α2] = [β − ε] ≥ 2. We have nβ − 1 ≤ [nβ − ε] < nβ − ε, and therefore
[(n+ 1)β − ε] − [nβ − ε] < (n+ 1)β − ε − nβ + 1 = β + 1 − ε ≤ k + 2 − ε,
[(n+ 1)β − ε] − [nβ − ε] > (n+ 1)β − 1 − nβ + ε = β − 1 + ε > k − 1 + 2ε.
Therefore
k ≤ [(n+ 1)β − ε] − [nβ − ε] = [(n+ 1)α2] − [nα2] ≤ k + 1.
The difference between two consecutive terms of the sequence [nβ − ε] is equal tothe difference between consecutive missing terms from the sequence [nα1]; that is, kor k+ 1 is the same as the difference between two consecutive terms of the sequence[nα2]. This implies the fact that the j th term that does not belong to S1 is precisely[jα2] = [jβ − ε]. This implies that m = 2.
If α1 ∈ Q, then α2 = b/d, and for j = d, we obtain [jα2] > [jβ − ε], which isfalse.
6.2 Algebraic Combinatorics 101
Comments 40 The fact that the two sequences from the previous problem are com-plementary is known as the Rayleigh–Beatty theorem, since S. Beatty proposed it asa problem in Amer. Math. Monthly in 1926. Presumably, this was known to Rayleigh,who mentioned it without proof in 1894. In 1927, J.V. Uspensky proved that if the msequences are complementary, thenm ≤ 2; his proof was simplified in several papersby Skolem, Graham, and Fraenkel, who also provided a far-reaching generalizationin 1969.
• S. Beatty: Problem 3173, Amer. Math. Monthly 33 (1926), 3, 156.• J. Lambek, L. Moser: Inverse and complementary sequences of natural numbers,
Amer. Math. Monthly 61 (1954), 454–458.• A.S. Fraenkel: Complementary systems of integers, Amer. Math. Monthly 84
(1977), 114–115.• A.S. Fraenkel: The bracket function and complementary sets of integers, Canad.
J. Math. 21 (1969), 6–27.• R.L. Graham: On a theorem of Uspensky, Amer. Math. Monthly 70 (1963), 407–
409.• Th. Skolem: On certain distributions of integers in pairs with given differences,
Math. Scand. 5 (1957), 57–68.• J.V. Uspensky, M.A. Heaslet: Elementary Number Theory, McGraw-Hill, New
York, 1939.
Problem 2.28. Let f : Z+ → Z+ be an increasing function and set
F(n) = f (n)+ n, G(n) = f ∗(n)+ n,
where f ∗(n) = card({x ∈ Z+; 0 ≤ f (x) < n}). Then {F(n); n ∈ Z+} and{G(n); n ∈ Z+} are complementary sequences. Conversely, any two complemen-tary sequences can be obtained this way using some nondecreasing function f .
Solution 2.28. We compute the number N of integers from {F(n),G(n), n ∈ Z+}that are smaller than M . Suppose that k terms of the form G(n) are smaller than M .Then
f ∗(k)+ k < M ≤ f ∗(k + 1)+ k + 1,
and so f ∗(k) < M − k < f ∗(k + 1) + 1. Since f ∗(k) = card{x ∈ Z+; f (x) <k} < M − k, we derive f (M − k) ≥ k. Further, f ∗(k + 1) = {x ∈ Z+; f (x) <k + 1} ≥ M − k − 1 implies that f (M − k − 1) < k + 1. Therefore F(M − k) =f (M − k)+M − k ≥ M and F(M − k − 1) < M . This means that exactly M − k
terms of the form F(n) are smaller than M; therefore N = M .The converse is immediate by defining f (n) = F(n)− n.
Comments 41 Since f ∗ is nondecreasing, it follows that it makes sense to definef ∗∗. The sequencesG(n) andH(n), whereH(n) = f ∗∗(n)+n, are complementary,and thus f ∗∗ = f .
The result is due to A. Frankel.
102 6 Algebra and Combinatorics Solutions
Problem 2.29. LetM denote the set of bijective functions f : Z+ → Z+. Prove thatthere is no bijective function between M and Z.
Solution 2.29. We will find an injection ρ : R \ (Q ∪ [0, 2]) → M , as follows. Toany α ∈ R \ (Q ∪ [0, 2]) there is associated an irrational β < 2 such that 1
α+ 1
β= 1.
Let ρ(α) : Z+ → Z+ be the function defined by
ρ(α)(n) ={[α n2
], for even n,[
β n+12
], for odd n.
According to the previous problem, ρ(α) : Z+ → Z+ is a bijection. Moreover, themap ρ is easily seen to be injective. Thus card(M) ≥ card(R \ Q) = card(R) >card(Z), and the result follows.
Problem 2.30. Let F ⊂ Z be a finite set of integers satisfying the following proper-ties:
1. For any x ∈ F , there exist y, z ∈ F such that x = y + z.2. There exists n such that, for any natural number 1 ≤ k ≤ n, and any choice ofx1, . . . , xk ∈ F , their sum x1 + · · · + xk is nonzero.
Prove that card(F ) ≥ 2n+ 2.
Solution 2.30. By hypothesis, 0 �∈ F . Let F+ = F ∩ Z+, F− = F ∩ Z−, so thatF = F+ ∪ F−. Consider the unoriented graph � whose vertices are the elements ofF+ and whose edges are defined below: x and y are adjacent if there exists z ∈ F
such that x = y + z. By the first hypothesis, each vertex is adjacent to at least oneedge. This implies that the graph � contains a cycle [x1, . . . , xk]. Assume that k isminimal with this property. This means that there exist zi ∈ F such that:
x1 = x2 + z1 = x3 + z2 + z1 = · · · = x1 + zk + · · · + z1,
which implies thatz1 + · · · + zk = 0.
The second hypothesis implies that k ≥ n + 1, and thus card(F+) ≥ k ≥ n + 1. Asimilar argument shows that card(F−) ≥ n+ 1, and the claim follows.
Problem 2.31. For a finite graphGwe denote byZ(G) the minimal number of colorsneeded to color all its vertices such that adjacent vertices have different colors. Thisis also called the chromatic number of G.
Prove that the inequality
Z(G) ≥ p2
p2 − 2q
holds if G has p vertices and q edges.
6.2 Algebraic Combinatorics 103
Solution 2.31. If we fixN = χ(G) and the number p of vertices of the graphG, thenwe have to show that the number q of edges allowed
q ≤ p2
2
(1 − 1
N
).
By hypothesis, there exists a partition (according to the colors) of the set of vertices intoN classes such that two vertices in the same class are not adjacent. Let n1, n2, . . . , nNbe the number of vertices in the respective classes. The only possibility to get an edgeis to join two vertices lying in different classes, and thus the total number q of edgesis at most
∑1≤i<j≤N ninj . Further, one knows that
∑1≤i≤N ni = p. We now have
∑
i<j
ninj ≤ C2N
p2
N2 = p2
2
(1 − 1
N
).
The equality holds if and only if N divides p and ni = p/N , for all i.
Comments 42 It can be proved, in the same way, that we have
χ(G) ≥ p
[t]
(1 − t − [t]
1 + [t]
),
where t = p − 2qp
. There exists also an upper bound for an arbitrary graph, namely
χ(G) ≤ 1
2
(1 + √
8q + 1).
One problem that received a great deal of attention in the past was to get bounds forthe chromatic number of a graph in terms of its topology. For instance, assume thatthe graph G is drawn on some surface so that its edges are not crossing each other,in which case the graph is said to be embedded. If the surface is partitioned intocurvilinear polygons (called “countries”), then the dual graph of this decompositionis constructed by associating a vertex to each country and joining two vertices ifthe respective countries have a common frontier. This way one constructs all graphsembedded in that surface. Surfaces in R
3 are characterized by only one number,called the genus g, or equivalently, by its Euler–Poincaré characteristic, which is2 − 2g. The latter is computed elementarily starting from an arbitrary partition intocurvilinear polygons. If such a partition consists of V vertices, E edges, and Fpolygonal countries, then
2 − 2g = V − E + F.
Computing a bound for the chromatic number of an arbitrary graph embedded in asurface is equivalent to finding the number of colors needed for coloring any mapof the surface so that adjacent countries have different colors. Now, if G is a graphthat can be embedded in a surface of genus g, then it was stated without proof byHeawood in 1890 that
104 6 Algebra and Combinatorics Solutions
χ(G) ≤[
1
2(7 + √
1 + 48g)
].
This has been known as the Heawood conjecture since then, and it was finally proved in1968 for all surfaces except for the plane by Ringel and Youngs. The case of the planeremained open until 1977, when Appel and Haken presented a computer-assistedproof of that case, thenceforth known as the “four color theorem.” The article ofStahl presents the history of the problem and of the proofs.
• K. Appel, W. Haken: Every Planar Map is Four Colorable, With the collaborationof J. Koch, Contemporary Mathematics, 98, American Mathematical Society,Providence, RI, 1989.
• G. Ringel: Map Color Theorem, Die Grundlehren der mathematischen Wis-senschaften, Band 209, Springer-Verlag, 1974.
• S. Stahl: The other map coloring theorem, Math. Magazine 58 (1985), 3, 131–145.
Problem 2.32. LetDk be a collection of subsets of the set {1, . . . , n} with the propertythat whenever A �= B ∈ Dk , then card(A ∩ B) ≤ k, where 0 ≤ k ≤ n − 1. Provethat
card(Dk) ≤ C0n + C1
n + C2n + · · · + Ck+1
n .
Solution 2.32. LetPk be the set of those subsets of {1, . . . , n} with at most k elements.Then any maximal (with respect to inclusion) collection Dk as above must containPk , since otherwise, Dk ∪ Pk will still satisfy the requirements while being strictlylarger than Dk . Therefore, let us consider a maximal Dk and set Dk = Dk \ Pk . ForA ∈ Dk , let us set D = {A1, . . . , As | card(Ai) = k + 1, and Ai ⊂ A}. If B ∈ Dkand A ∩ B �= ∅, then there exist i, j such that Ai = Bj , where card(Ai) = k + 1;this would imply that A ∩ B ⊃ Ai and thus card(A ∩ B) ≥ k + 1, which is absurd.
Therefore, the sets A, for A ∈ Dk , are disjoint. Since each A has at least oneelement and A ⊂ Pk+1 \ Pk , we obtain that card(Dk) ≤ card (Pk+1 \ Pk) and hencecard(Dk) ≤ card(Dk)+ card(Pk) = card(Pk+1) = C0
n +C1n +C2
n +· · ·+Ck+1n . We
have equality for Dk = Pk+1.
Problem 2.33. Prove that
1
p!
n∑
k=0
(−1)n−kCknkp =
⎧⎪⎪⎪⎪⎪⎪⎪⎨
⎪⎪⎪⎪⎪⎪⎪⎩
0, if 0 ≤ p < n,
1, if p = n,
n/2, if p = n+ 1,n(3n+1)
24 , if p = n+ 2,n2(n+1)
48 , if p = n+ 3,n(15n3+30n2+5n+1)
1152 , if p = n+ 4.
Solution 2.33. Consider the function (ex−1)n, in which we develop first the binomialand then we develop the exponentials in Taylor series:
6.2 Algebraic Combinatorics 105
(ex − 1)n =n∑
k=0
(−1)n−kCknekx =n∑
k=0
(−1)n−kCkn( ∞∑
j=0
1
j !kjxj
)
=∞∑
j=0
1
j !
( n∑
k=0
(−1)n−kCknkj).
For the same function we now develop the exponential and then the binomial asfollows:
(ex − 1)n =(x + 1
2!x2 + 1
3!x3 + · · ·
)n
= xn + n
2xn+1 + n(3n+ 1)
24xn+2 + n2(n+ 1)
48xn+3
+ n(15n3 + 30n2 + 5n+ 1)
1152xn+4 + · · · .
By identifying the first coefficients in the two series, we obtain the formulas from thestatement.
Comments 43 The problem of computing the sums sn,p = ∑nk=0(−1)kCknk
p is aclassical one, and the first result (for p = n) is due to Leonhard Euler. D. Andricagave a method of computation by means of the recurrence relations
sn,p+1 = n(sn,p − sn−1,p), p ≥ 0.
In particular, one recovers the results from the problem.
• D. Andrica: On a combinatorial sum, Gazeta Mat. 5 (1989), 158.
Problem 2.34. Write lcm(a1, . . . , an) in terms of the various gcd(ai, . . . , aj ) forsubsets of {a1, . . . , an}.Solution 2.34. Let Pj = ∏
1≤i1<···<ij≤n gcd(ai1 , ..., aij ). We have
lcm(a1, . . . , an) = P1P3P5 · · ·P2P4P6 · · · .
For instance, if n = 3 we have
lcm(x, y, z) = xyz gcd(x, y, z)
gcd(x, y)gcd(y, z)gcd(z, x).
Let p be a prime and ei maximal such that pei divide ai . We can assume that en ≤en−1 ≤ . . . ≤ e1. In Pj , the prime p appears to the ei th power in Cj−1
n−1 cases, andso the exponent of p in
∏i P2i+1/
∏i P2i is e1, because the alternating sum of the
binomial coefficients is zero, with the exception of C00 = 1.
106 6 Algebra and Combinatorics Solutions
Problem 2.35. Let f (n) be the number of ways in which a convex polygon with n+1sides can be divided into regions delimited by several diagonals that do not intersect(except possibly at their endpoints). We consider as distinct the dissection in whichwe first cut the diagonal a and next the diagonal b from the dissection in which wefirst cut the diagonal b and next the diagonal a. It is easy to compute the first valuesof f (n), as follows: f (1) = 1, f (2) = 1, f (3) = 3, f (4) = 11, f (5) = 45. Find thegenerating function F(x) = ∑
f (n)xn and an asymptotic formula for f (n).
Solution 2.35. By a simple counting argument, we obtain the recurrence
f (n) = 3f (n− 1)+ 2n−2∑
k=2
f (k)f (n− k).
If we write y = F(x)− x, then y satisfies the functional equation y = x2 + 3xy +2y2. This is a consequence of the recurrence relation above. Solving the degree-twoequation in y, we derive that
F(x) = 1
4
(1 + x −
√1 − 6x + x2
).
Using Taylor’s formula and induction on n, one further shows that
f (n) =√
3√
2 − 4
4√π
·(
3 + 2√
2)n
n√n
(
1 + 3(8 − 3√
2)
32n+ O
(n−2
))
.
Problem 2.36. Find the permutation σ : (1, . . . , n) → (1, . . . , n) such that
S(σ) =n∑
i=1
|σ(i)− i|
is maximal.
Solution 2.36. We want to prove that a permutation σ that maximizes S(σ) satisfiesσ(1) > σ(2) > · · · > σ(n).
Let us prove first the following intermediate result. If a < c, b < d, then theinequality
|a − b| + |c − d| ≤ |a − d| + |b − c|holds with equality when either a < c < b < d or b < d < a < c.
From the symmetry of the previous inequality, we can assume that a < b, d,and using a translation followed by a homothety on the real axis, we can, moreover,assume that a = 0, c = 1.
If b < 1, d < 1, the inequality reduces to 2b < 2d.If b < 1, d > 1, then the inequality is 2b < 2.If b > 1, d > 1, we obtain equality. This proves the claim.
6.2 Algebraic Combinatorics 107
Now let σ be a permutation such that S(σ) is maximal. Suppose that there existsi < j such that σ(i) < σ(j). Let us define another permutation σ ∗ = σ ◦ (i, j),where (i, j) denotes the transposition interchanging i and j .
According to our claim, we have S(σ ∗) ≥ S(σ).By modifying iteratively σ by taking the product with the transpositions deter-
mined by all pairs (i, j) as above, we will finally obtain γ :
γ :
(1 2 · · · nn n− 1 · · · 1
).
Moreover, the inequality above shows that
S(σ) ≤ S(γ ) =n∑
i=1
|n− 2i + 1|.
Problem 2.37. On the set Sn of permutations of {1, . . . , n} we define an invariantdistance function by means of the formula
d(σ, τ ) =n∑
i=1
|σ(i)− τ(i)|.
What are the values that d could possibly take?
Solution 2.37. We have d(ρσ, ρτ) = d(σ, τ ), for any ρ, σ, τ ∈ Sn. Therefore, itsuffices to consider the values of d(1, σ ), where 1 is the identity permutation.
If m > 0 is a value of d, we will show below that m− 2 is also a value of d, andhence d takes all even values from 0 to some 2t , where t has to be determined.
Notice first that d takes only even values, because
d(σ, τ ) ≡n∑
i=1
(σ (i)− τ(i)) ≡n∑
i=1
σ(i)−n∑
i=1
τ(i) ≡ 0 (mod 2).
Letm = d(1, σ ) > 0. There then exist 1 ≤ r < s ≤ n such that σ(r) > r, σ (s) <
s, and σ(i) = i, if r < i < s.Let ρr,s be the cycle (r, r + 1, . . . , s). We claim that d(1, ρr,sσ ) = m− 2. First,
by hypothesis, σ(r) > s and σ(s) < r . Looking at the contribution of the elementsfrom r to s to the respective distances, we obtain
d(1, σ )−d(1, ρr,sσ ) = (σ (r)−r+s−σ(s))−(r−σ(s)+s−r−1+σ(r)−r−1) = 2.
Let us show now that t =[n2
4
](see also the previous problem) and the maximum
distance d(1, σ ) is realized for the permutation θ(i) = n+ 1 − i.
We have d(1, θ) = 2[n2
4
]=
[n2
2
]. Let k < n−1, such that σ(i) = n+1− i, for
all i < k, while σ(k) �= n+1−k. We claim that there exists some permutation σ ∈ Snwith the property that d(1, σ ) ≤ d(1, σ ) and that satisfies σ (i) = n + 1 − i, i ≤ k.Then, using induction k, we find a sequence of permutations reaching θ such that
108 6 Algebra and Combinatorics Solutions
d(1, σ ) ≤ d(1, σ ) ≤ · · · ≤ d(1, θ).
If σ(k) �= n+ 1 − k, we have σ(k) < n+ 1 − k because every number greater thann + 1 − k is the image of some i < k. But this implies that σ(r) = n + 1 − k, forsome r > k. Let τ be the transposition (k, r). Set σ = τσ ; we have
d(1, σ )− d(1, σ ) = |r − kσ | + |m+ 1 − 2k| − |k − kσ | + |n+ 1 − k − r| = w.
Let us suppose that k ≤ σ(k) ≤ r ≤ n + 1 − k (the other cases are similar). Thenw = r−σ(k)+n+1−2k−σ(k)+k−n−1+k+r = 2(r−kσ) ≥ 0. Analogously,for all k, r we have
d(1, σ )− d(1, σ ) ≥ 0,
which proves the claim.
Problem 2.38. The set M = {1, 2, . . . , 2n} is partitioned into k sets M1, . . . ,Mk ,where n ≥ k3 + k. Show that there exist i, j ∈ {1, . . . , k} for which we can findk + 1 distinct even numbers 2r1, . . . , 2rk+1 ∈ Mi with the property that 2r1 −1, . . . , 2rk+1 − 1 ∈ Mj .
Solution 2.38. There exists a set Ms that contains at least 2nk
≥ 2(k2 + 1) elements.We have to consider two cases:
1. EitherMs contains at least 2(k2+1)2 = k2 +1 even numbers. Then the set of odd
numbersO = {r − 1,where r is even, and r ∈ Ms}
has k2 + 1 elements. Then there exists some setMa containing at least k2+1k
elementsfrom O. We choose therefore i = s and j = a. Notice that i might be equal to j .
2. Or elseMs contains at least k2 + 1 odd numbers. The solution is similar to thatfrom above, considering the set of even numbers.
Problem 2.39. Let S be the set of odd integers not divisible by 5 and smaller than30m, wherem ∈ Z
∗+. Find the smallest k such that every subset A ⊂ S of k elementscontains two distinct integers, one of which divides the other.
Solution 2.39. Consider the subset N = {1, 7, 11, 13, 17, 19, 23, 29, . . . , 30m − 1}of elements of S that are not divisible by 3. There are 8m elements in N , which arewritten in increasing order a1 < a2 < · · · < a8m. Every element of S can be uniquelywritten as x = ai · 3t , where t ∈ Z+ and ai ∈ N .
If k ≥ 8m+ 1, then according to the pigeonhole principle, any subset A ⊂ S ofcardinality card A = k contains two distinct elements x, y ∈ A for which x = ai3t
and y = ai3q . In this case, either x divides y, or y divides x.Next, for all i, choose the maximal t (i) with the property that ai3t (i) < 30m <
ai3t (i)+1, and set bi = ai3t (i). We have therefore 10m < bi < 30m. The set{b1, b2, . . . , b8m} contains 8m elements from S, and also we have the inequalities0 < bi/bj < 3, for any i, j . Since all numbers bi are odd, we derive that bi/bj �∈ Z.Therefore {b1, . . . , b8m} does not contain a number and one divisor of it. This showsthat the required value of k is 8m+ 1.
6.2 Algebraic Combinatorics 109
Problem 2.40. Prove that∏
1≤j<i≤nai−aji−j is a natural number whenever a1 ≤ a2 ≤
· · · ≤ an are integers.
Solution 2.40. We can suppose ai ≥ 0. Then,
S =∏
i>j
ai − aj
i − j=
∏i>j (ai − aj )
(n− 1)!(n− 2)! · · · 1!.
We now consider the determinant D of the matrix⎛
⎜⎜⎜⎝
1 1 · · · 1C1a1
C1a2
· · · C1an
......
...
Cn−1a1
Cn−1a2
· · · Cn−1an
⎞
⎟⎟⎟⎠.
It is obvious thatD ∈ Z+. We now write the binomial coefficientCkm using factorials,and we arrange the factors in the lines. We obtain
D = 1
1!2! · · · (n− 1)!det
⎛
⎜⎜⎜⎝
1 1 · · · 1a1 a2 · · · an...
......
an−11 an−1
2 · · · an−1n
⎞
⎟⎟⎟⎠
=∏i>j ai − aj
∏i=1(i − 1)!
= S,
and the claim follows.
Problem 2.41. Is there an infinite setA ⊂ Z+ such that for all x, y ∈ A neither x norx + y is a perfect power, i.e., ak , for k ≥ 2? More generally, is there an infinite setA ⊂ Z+ such that for any nonempty finite collection xi ∈ A, i ∈ J , the sum
∑i∈J xi
is not a perfect power?
Solution 2.41. 1. Yes. Let p1 = 2, p2 = 3, . . . , pn, . . . be the set of primes in in-creasing order. We then set A = {223, 22325, . . . , 223252 · · ·p2
npn+1, . . .}.Obviously, no element of A is a perfect power. Moreover, if x ≤ y ∈ A, then we
can write x = 22 · · ·p2kpk+1 and x = 22 · · ·p2
npn+1. If k < n, then pk+1 dividesx+y = 22 · · ·p2
kpk+1(1+pk+1p
2k+2 · · ·p2
npn+1), while p2
k+1 does not divide x+y,and thus it cannot be a perfect power. If k = n, the same argument works, unless k = 0,which was excluded from A.
2. Let us prove that if C ⊂ Z+ is a set of density d(C) = 0, then there exists aninfinite set A ⊂ Z+ such that for any nonempty finite collection xi ∈ A, i ∈ J , thesum
∑i∈J xi is not in C.
Recall that the density of the subset C ⊂ Z+ is defined as
d(C) = limn→∞
card(C ∩ {1, 2, . . . , n})n
.
Since d(C) = 0, there exists a natural number a1 �∈ C. Then consider C1 =C ∪ (C − a1) ∩ Z+, where we set X − λ = {x − λ|x ∈ X}. The density d has thefollowing fundamental property that results from the definition:
110 6 Algebra and Combinatorics Solutions
d(A ∪ B) ≤ d(A)+ d(B).
This implies that d(C1) = 0, and therefore there exists a natural number a2 �∈ C1.Assuming that both the subsetCk of density zero and the sequencea1, a2, . . . , ak+1,
with ak+1 �∈ Ck , are defined, we set
Ck+1 = Ck
k⋃
p=1
⋃
i1≤i2≤···≤ip≤k+1
(Ck − (ai1 + ai2 + · · · + aip )) ∩ Z+.
It follows that d(Ck+1) = 0 and thus there exists at least one integer ak+2 > ak+1such that ak+2 �∈ Ck+1.
The sequence (ak) forms an infinite subset A with the required property.It suffices now to compute the density of the set C = {ak|a ≥ 2, k ≥ 2}. We have
card(C ∩ {1, 2, . . . , n})n
= [√n] + [ 3
√n] + · · ·
n≤
√n log2 n
n= log2 n√
n.
Therefore d(C) = 0.
Comments 44 There is an analogous result when products are used instead of sumsof elements in A. For instance, the set
A ={a1, a2 = C
a12a1, C
a2a1+a2
, . . . , Canan−1+an
}
has the property that no product of its elements is a perfect power. Is it true thatd(A) = 0?
Problem 2.42. Let
f (n) = maxAAA
...
Ak
32
1 ,
where n = A1 +· · ·+Ak . Thus, f (1) = 1, f (2) = 2, f (3) = 3, f (4) = 4, f (5) =9, f (6) = 27, f (7) = 512, etc. Determine f (n).
Solution 2.42. It is clear that if f (n) = A···Ak
1 , then Ai > 1. Also, f is an increasing
function, and thus f (n+ 1) ≥ f (n)+ 1. If A1 = k, then A···Ak
2 ≤ f (n− k), and thisimplies that whenever n ≥ 4, we have
f (n) = max2≤k≤n−2
kf (n−k).
It is easy now to see that f (n + 1) ≥ 2f (n). In fact, let f (n) = kf (n−k), for somek ≥ 2; then
f (n+ 1) ≥ kf (n−k+1) ≥ kf (n−k)+1 ≥ 2f (n).
The next step is to compare af (b+1) with (a + 1)f (b), that is, f (b+1)f (b)
with log(a+1)log(a) .
If a ≥ 2, then a + 1 < a2 and hence log(a + 1)/ log a < 2. Thus, if b ≥ 4, then
6.2 Algebraic Combinatorics 111
f (b + 1)/f (b) ≥ 2, which yields af (b+1) > (a + 1)f (b), as soon as a ≥ 2, b ≥ 3.For small values of the arguments, we have
log 4
log 3<f (4)
f (3)<f (3)
f (2)<
log 3
log 2.
These show that kf (n−k) has a maximum when k = 2 if n > 6. Therefore, the finalanswer is
f (n) =
⎧⎪⎨
⎪⎩
2···232
, for odd n > 3,
2···233
, for even n > 4.
Problem 2.43. Consider a setM withm elements andA1, . . . , An distinct subsets ofM such that card(Ai ∩ Aj) = r ≥ 1 for all 1 ≤ i �= j ≤ n. Prove that n ≤ m.
Solution 2.43. Observe first that ifA is anm×nmatrix andm > n, then det(A·A�) =
0, where A� denotes the transpose matrix. In fact, let us border the matrix A by anull matrix of size (m− n)× n, in order to get anm×mmatrix A′. It is obvious thatA′ · A′� = A · A�, and thus det
(A · A�) = det
(A′ · A′�) = 0.
Let D be the n× n matrix whose entries are
Dij ={r ≥ 1, if i �= j,
dii ≥ r, otherwise.
Assume that there exists at most one i such that dii = r . We claim then that det(D) �=0. This follows by subtracting the last line from the first n − 1 and an inductiveargument.
Consider now the m× n matrix A whose entries are
Aij ={
1, if i ∈ Aj ,0, otherwise.
It is now immediate that D = A · A� is given by
Dij =n∑
j=1
aikakj = card(Ai ∩ Aj).
By hypothesis,Dij = r when i �= j , and since theAi are distinct, there exists at mostone i such that Dii = r .
The claim from above tells us that det(D) = det(A · A�) �= 0, while the first
observation implies that m ≤ n.
Problem 2.44. Setπ(n) for the number of prime numbers less than or equal ton. Provethat there are at most π(n) numbers 1 < a1 < · · · < ak ≤ n with gcd(ai, aj ) = 1.
112 6 Algebra and Combinatorics Solutions
Solution 2.44. If X is a set, we denote by P(X) the set of nonempty subsets of X.Define the map h : {1, . . . , n} → P({p1, . . . , pπ(n)}) that associates to the integer xthe set of prime divisors of x.
One then has card(h(ai)) ≥ 1 and furtherh(ai)∩h(aj ) = ∅, from our assumption.Since the maximal number of disjoint subsets of {p1, . . . , pπ(n)} is bounded by π(n),the claim follows.
Problem 2.45. Prove that for every k, there exists n such that the nth term of theFibonacci sequenceFn is divisible by k. Recall thatFn is determined by the recurrenceFn+2 = Fn + Fn+1, for n ≥ 0, where the first terms are F0 = 0, F1 = 1.
Solution 2.45. Using the pigeonhole principle, there exists an infinite sequence nisuch that
Fn1 ≡ Fn2 ≡ · · · ≡ Fnm ≡ · · · (mod k).
Moreover, there exists an infinite subsequence nis such that
Fni1+1 ≡ Fni2 +1 ≡ · · · ≡ Fnim+1 ≡ · · · (mod k)
by the same argument. Thus one knows that
Fnis ≡ Fnit (mod k) and Fnis+1 ≡ Fnit+1 (mod k).
The recurrence relation of the Fibonacci numbers shows that
Fnis+m ≡ Fnit+m (mod k), for any m ∈ Z.
Since F0 = 0, we find that there exists n (actually infinitely many such integers) suchthat k divides Fn.
Problem 2.46. Consider a set of consecutive integers C + 1, C + 2, . . . , C + n,where C > nn−1. Show that there exist distinct prime numbers p1, p2, . . . , pn suchthat C + j is divisible by pj .
Solution 2.46. Let k ≤ h and consider the factorization of C + k into prime factors.If C+ k has at least n distinct prime divisors, then we can find one prime divisor thatis not yet associated with the other n − 1 numbers C + i. Assume then that C + k
has at most j ≤ n− 1 prime factors, which are p1k, . . . , pjk for 1 ≤ j ≤ n− 1. Wewrite down the factorization as
pa1k1k · · ·pajkjk = C + k > C > nn−1.
The pigeonhole principle implies that there exists some prime power qk = pajkjk such
that qk > n.We then associate the prime number pjk toC+k. Let us show that this procedure
yields distinct prime numbers. Suppose that there exist j and k such that the associatednumbers are the same. We know that qk divides C + k and qj divides C + j , andthey are both prime powers of the same prime. Since qk > n and qj > n, we obtaingcd(qk, qj ) > n. Moreover, gcd(qk, qj ) divides both C + k and C + j and hencetheir difference |j − k| < n, which is a contradiction. This proves the claim.
6.2 Algebraic Combinatorics 113
Problem 2.47. Let p be a prime number and f (p) the smallest integer for whichthere exists a partition of the set {2, 3, . . . , p} into f (p) classes such that whenevera1, . . . , ak belong to the same class of the partition, the equation
k∑
i=1
xiai = p
does not have solutions in nonnegative integers. Estimate f (p).
Solution 2.47. 1. Suppose that a1, a2 are in the same class and gcd(a1, a2) = 1. Oneknows then that any natural number greater than or equal to a1a2 can be written asx1a1 + x2a2, where xi ∈ Z+. Therefore, the prime numbers smaller than
√p should
to different classes of our partition, and hence
f (p) > π(p
12) ∼ 2
√p
logp,
where π(n) denotes the number of primes smaller than n.2. If t ∈ Z+, let us consider the sets of consecutive elements At ={[pt+1
]+ 1, . . . ,
[pt
]}. We claim that the equation
∑ki=1 aixi = p has no integer
solutions if ai ∈ At for 1 ≤ i ≤ k. In fact, if we had a solution xi , then we wouldhave
p
t + 1
k∑
i=1
xi <
k∑
i=1
aixi <p
t
k∑
i=1
xi,
and so t <∑ki=1 xi < t + 1, which is false, since xi are integers.
Let us consider the classes A1, A2, . . . , AL−1, where L is an integer to be fixedlater.
Consider now, for every prime q < p/L, the classesBq = {q, 2q, . . . , αq, . . .}. Itis immediate that the associated linear equation has no solutions if the coefficients be-long to someBq , since q does not dividep. Further,A1, . . . , AL−1, B2, B3, . . . , B
[ pL
]
form a partition of {2, 3, . . . , p}. The total number of classes of this partition is thenL− 1 + π(p/L).
Setting L =√
2plogp , we obtain f (p) <
√8p
logp (1 + O(1)).
Problem 2.48. Consider m distinct natural numbers ai smaller than N such that
lcm(ai, aj ) ≤ N for all i, j . Prove that m ≤ 2[√N].
Solution 2.48. Assume that the numbers ai are ordered as 1 ≤ am < · · · < a1 ≤ N .We will prove by induction on k that ak ≤ N
k. If k = 1, then it is obvious. Moreover,
if k ≥ 1, then
ak − ak+1 ≥ gcd(ak, ak+1) = akak+1
lcm(ak, ak+1)≥ akak+1
N.
This is equivalent to ak+1 + akak+1N
≤ ak , which yields
114 6 Algebra and Combinatorics Solutions
ak+1 ≤ Nak
ak +N= N − N2
ak +N≤ N − N2
Nk
+N= N
k + 1,
and our claim follows. Furthermore, one has
a[√N]+1
≤ N[√N] + 1
≤ [√N].
The sequence (aj ) contains then at most [√N ] terms between 1 and
[√N]. On the
other hand, between√N andN , there are no more than
[√N]
terms of our sequence,and thus we have a maximum number of m ≤ 2
[√N]
terms.
Comments 45 If m(N) denotes the maximum number of terms of a sequence (ai),as in the statement, then Erdos conjectured the following asymptotic behavior of thefunction m(n):
m(N) = 3
23/2
√N + O(1).
See also:
1. P. Erdos: Remarks on number theory. IV. Extremal problems in number theory,Matematikai Lapok 13 (1962), 228–255.
Problem 2.49. The setM ⊂ Z+ is calledA-sum-free, whereA = (a1, a2, . . . , ak) ∈Zk+, if for any choice of x1, x2, . . . , xk ∈ M we have a1x1 + a2x2 + · · · + akxk �∈ M .
IfA,B are two vectors, we define f (n;A,B) as the greatest number h such that thereexists a partition of the set of consecutive integers {n, n + 1, . . . , h} into S1 and S2such that S1 is A-sum-free and S2 is B-sum-free. Assume that B = (b1, b2, . . . , bm)
and that the conditions below are satisfied:
a1 + a2 + · · · + ak = b1 + b2 + · · · + bm = s,
andmin
1≤j≤k aj = min1≤j≤m bj = 1, k,m ≥ 2.
Prove that f (n;A,B) = ns2 + n(s − 1)− 1.
Solution 2.49. First consider the sets
S1 = {n, n+ 1, . . . , ns − 1} ∪ {ns2, ns2 + 1, . . . , ns2 + n(s − 1)− 1},S2 = {ns, ns + 1, . . . , ns2 − 1}.
If xj ∈ S2, then b1x1 + b2x2 + · · · + bmxm ≥ ns2 and thus S2 is B-sum-free.If x1, x2, . . . , xk ∈ S1, then we have two cases:
1. If xi ≤ ns − 1 for all i, then
ns ≤ a1x1 + a2x2 + · · · + akxk ≤ s(ns − 1) < ns2.
6.2 Algebraic Combinatorics 115
2. If some xi belongs to {ns2, ns2 + 1, . . . , h}, then
a1x1 + a2x2 + · · · + akxk ≥ ns2 + n(s − 1).
Thus S1 is A-sum-free. This proves that f (n;A,B) ≥ ns2 + n(s − 1)− 1.Assume now that the set {n, n + 1, . . . , ns2 + n(s − 1)} can be partitioned into
two sets Si that are A (respectively B) sum-free.Let n ∈ S1. If we take x1 = x2 = · · · = xk = n, then ns = a1x1 + a2x2 + · · · +
akxk �∈ S1, so ns ∈ S2. Taking now x1 = x2 = · · · + = xm = ns ∈ S2, we find thatns2 ∈ S1.
Suppose that a1 = b1 = 1. If we consider the elements x1 = ns2, x2 = · · · =xk = n of S1, we derive that n(s2 + s − 1) ∈ S2.
1. If n(s+1) ∈ S1, then take x1 = n, x2 = · · · = xk = n(s+1), and since ns2 ∈ S1,we find that the set S1 is not A-sum-free.
2. If n(s + 1) ∈ S2, then take x1 = ns, x2 = · · · = xm = n(s + 1), and sincen(s2 + s − 1) ∈ S2, we obtain that the set S2 is not B-sum-free.
This contradiction shows that f (n;A,B) = ns2 + n(s − 1)− 1.
Comments 46 Generalized sum-free sets of integers were considered first by Rado,who in 1933 gave the upper bound
f (1;A,A) ≤ max
((bmc2 − 1)(c − 1)+ bmc,
bmc2(c − 1)
a
),
where A = (a, b), m = agcd(a,b) and c = max(x0, y0, z0), where (x0, y0, z0) is the
minimal solution of the Diophantine equation ax+ by = z. There are several partialresults known for particular forms of A and B. If we denote the vector (d, d, . . . , d)with k components by k〈d〉, then Kasá proved that
f (1; 2〈1〉, k〈1〉) ={
3k − 3, for odd k,3k − 2, for even k,
f (n; 2〈1〉, k〈1〉) = (2k + 1)n− 1, for even n.
Further, Seress improved this by showing that
f (n;m〈1〉, k〈1〉) = (mk +m− 1)n− 1, for n > 1,m ≥ 3.
The result from the problem is due to L. Funar, who proved also that
f (n; k〈d〉, k〈d〉) = k2d3 + kd − d − 1
for d even and k ≥ d and in several other cases. Abbott showed that this formulaholds for all k, d ≥ 2. Other estimates for f (n;A,B) in the case in whichA,B havenot necessarily the same sum of components have been obtained by P. Moree. Thesesuggest that there is no such simple closed formula for arbitrary A,B. However,recent progress has lead to the determination of f (1;A,A) for arbitrary A (also
116 6 Algebra and Combinatorics Solutions
known as the 2-color Rado number), by Guo and Sun (improving previous results byB. Hopkins and D. Schaal). Specifically, we have
f (1;A,A) = a(s − a)2 + (2a2 + 1)(s − a)+ a3,
where a = min1≤j≤k aj and s = a1 + · · · + ak .The general case, when we consider partitions into n ≥ 3 subsets Sj such that
each Sj is Aj -sum-free, seems to be much more difficult, and no general estimatesfrom above are known.
• H.L. Abbott: On a conjecture of Funar concerning generalized sum-free sets,Nieuw Arch. Wisk. (4) 9 (1991), 249–252.
• L. Funar: Generalized sum-free sets of integers, Nieuw Arch. Wisk. (4) 8 (1990),49–54.
• Song Guo and Zhi-Wei Sun: Determination of the two-color Rado number fora1x1 + · · · + amxm = x0, math.CO/0601409.
• B. Hopkins and D. Schaal: On Rado numbers for∑m−1i=1 aixi = xm, Adv. Appl.
Math. 35 (2005), 433–441.• P. Moree: On a conjecture of Funar, Nieuw Arch. Wisk. (4) 8 (1990), 55–60.• R. Rado: Studien zur Kombinatorik, Math. Zeitschrift 36(1933), 424–480.• A. Seress: k-sum-free decompositions, Matematikai Lapok 31 (1978/83), 191–
194.
Problem 2.50. Let 1 ≤ a1 < a2 < · · · < an < 2n be a sequence of natural numbersfor n ≥ 6. Prove that
mini,j
lcm(ai, aj ) ≤ 6([n
2
]+ 1
).
Moreover, the constant 6 is sharp.
Solution 2.50. We assume that among the ai’s we can find both a and 2a. Thenmini,j lcm(ai, aj ) ≤ lcm(a, 2a) ≤ 2n < 6
([n2
] + 1). Let a ≤ n be an element of
the sequence, if it exists. If 2a also belongs to the sequence, the claim follows fromabove. If 2a is not in the sequence, then we will replace the element a with 2a. Thisway the value of mini,j lcm(ai, aj ) is not diminished. We continue this process as faras possible. At some point, all the elements of the sequence will be greater than n,and thus the last sequence is forced to be n+ 1, n+ 2, . . . , 2n.
If n = 2k + 1, then take a = 2k + 2, b = 3k + 3, so that lcm(a, b) = 6k + 6 =6([n2
] + 1).
Now if c, d are such that n + 1 < c < d ≤ 2n, then we have lcm(c, d) ≥6([n2
] + 1). In fact, lcm(c, d) = pd = qc for some q > p > 1, and thus either
p = 2 and q ≥ 3, or else q ≥ 4.In the first case, c is an even number; hence c ≥ 2
([n2
] + 1)
and lcm(c, d) =qc ≥ 3c ≥ 6
([n2
] + 1).
In the second case, lcm(c, d) = qc ≥ 4(n + 1) > 6([n2
] + 1). Therefore, if we
choose the set n+ 1, . . . , 2n, we have mini,j lcm(ai, aj ) = 6([n2
] + 1).
6.2 Algebraic Combinatorics 117
Comments 47 This result is due to P. Erdos, who claimed also that under the samehypothesis, we have
maxi �=j gcd(ai, aj ) >
38n
147− C,
where C is a constant independent of n.
Problem 2.51. Let 1 ≤ a1 < a2 < · · · < ak < n be such that gcd(ai, aj ) �= 1 for all1 ≤ i < j ≤ k. Determine the maximum value of k.
Solution 2.51. Let f (n) denote the maximum of k as a function of n. We have thenf (2) = 1. Assume from now on that n ≥ 3.
Define li (for 1 ≤ i ≤ k) to be the smallest integer satisfying the inequalityn/2 < ai2li . If there exists a pair of distinct indices i and j such that ai2li = aj2lj ,then either ai divides aj , or conversely, aj divides ai . Therefore, we can replace aiby ai2li without diminishing the value of k.
Thus we can assume that ai ≥ n2 , for all i. Further, the condition gcd(ai, aj ) �= 1
tells us that we cannot find two consecutive numbers that both belong to the sequenceai . Thus, if n is of the form 4m− 1, 4m, or 4m+ 2, then f (n) ≤ m.
Moreover, if n = 4m + 1, then f (n) ≤ m + 1. If we have equality above, thenthe sequence ai is forced to be a1 = 2m+ 1, a2 = 2m+ 3, . . . , am+1 = 4m+ 1, andtherefore gcd(a1, a2) = 1, which is a contradiction. Thus f (n) ≤ m if n = 4m+ 1.
By considering the set of all even numbers in the interval [n/2, n] we obtain thatfor any n ≥ 3, we have f (n) = [
n+14
].
Problem 2.52. Consider the increasing sequence f (n) ∈ Z+, 0 < f (1) < f (2) <· · · < f (n) < · · · . It is known that the nth element in increasing order among thepositive integers that are not terms of this sequence is f (f (n))+ 1. Find the value off (240).
Solution 2.52. 1. One knows that the nth absent number is N = f (f (n))+ 1. How-ever, the numbers smaller than N that belong to the sequence are f (1), f (2), . . . ,f (f (n)). Moreover, there are n numbers less than or equal to N that do not belongto the sequence. This means that N = f (n)+ n; thus
f (f (n)) = f (n)+ n− 1.
Therefore f (1) = 1, f (2) = 3. It follows by induction that
f (n) = n+ card {m|f (m) < n}.Set a0 = 2, an+1 = f (an). We compute
an+1 = an+ card {m|f (m) < an} = an+ card {m|f (m) < an−1} = an+ an−1 − 1,
and therefore the sequence bn = an − 1 satisfies the Fibonacci recurrence bn+1 =bn + bn−1. We will prove by induction that
f (bn + x) = bn+1 + f (x), for 1 ≤ x ≤ bn−1.
118 6 Algebra and Combinatorics Solutions
If n = 0, then b1 = 1 and x = 1 and the claim is verified. Further
f (bn+ x) = bn+ x+ card {y|f (y) < f (bn−1)}+ card {y|bn+1 < f (y) < bn+ x},which implies that f (bn − 1) < f (an−1) = an and thus f (bn−1) ≤ bn ≤ bn + x. Inparticular,
f (y) ≤ bn + x ≤ bn + bn−1 ≤ bn+1 < f (an),
from which we obtain y < an, y ≤ bn. We now write
y = bn+1 + z, where z ≤ bn − bn−1 = bn−2.
Hence
f (bn + x) = bn + x + bn−1 + card {z|f (z) < x} = bn+1 + f (x).
This proves our claim.Recall now that every natural number can be uniquely written as a sum of
Fibonacci numbers, i.e., as
x = 1 + bk1 + bk2 + · · · + bkp ,
where 1 ≤ k1 < k2 < · · · < kp. If x is written as above, the previous inductiveformula yields the value
f (x) = 1 + bk1+1 + bk2+1 + · · · + bkp+1.
In particular,f (240) = 1 + 2 + 8 + 377 = 388,
since 240 = 1 + (1 + 5 + 233).2. We can use in a clever manner the recursion law f (f (n)) = f (n)+ n− 1 in
order to find, step by step,
f (3) = 3 + 1 = 4, f (4) = 4 + 2 = 6, f (6) = 6 + 3 = 9,
f (9) = 9 + 5 = 14, f (14) = 22
f (22) = 35, f (35) = 56, f (56) = 90, f (90) = 145,
f (145) = 234, f (234) = 378.
One knows that f (f (35))+ 1 = 91 is absent, so that f (57) = 92 and thus f (92) =148, f (148) = 239, f (239) = 386. Next f (f (148)) + 1 = 387 is absent, andhence f (240) = 388.
Problem 2.53. We define inductively three sequences of integers (an), (bn), (cn) asfollows:
1. a1 = 1, b1 = 2, c1 = 4;2. an is the smallest integer that does not belong to the set
{a1, . . . , an−1, b1, . . . , bn−1, c1, . . . , cn−1};
6.2 Algebraic Combinatorics 119
3. bn is the smallest integer that does not belong to the set{a1, . . . , an−1, an, b1, . . . , bn−1, c1, . . . , cn−1};
4. cn = 2bn + n− an.
Prove that0 < n
(1 + √
3)
− bn < 2 for all n ∈ Z+.
Solution 2.53. We will actually prove that
α < n(1 + √
3) − bn < β, where α = (
9 − 5√
3)/3, β = (
12 − 4√
3)/3.
Observe that the sequence (cn) does not contain two consecutive numbers. Thus an+1might jump at most two units ahead from bn, and bn two units ahead from an. Byinduction, we obtain
1 ≤ bn − an ≤ 2 and 1 ≤ an+1 − bn ≤ 2.
Using the equality
cn+1 − cn = 2(bn+1 − an+1)+ (an+1 − bn)− (bn − an)+ 1
we deduce that2 ≤ cn+1 − cn ≤ 6.
In particular, there are not six consecutive integers among the sequences aj and bj .Set γn = n
(1 + √
3)−bn. We will prove by induction that α < γn < β, which
is trivially verified for n = 1. Suppose now that this holds true for all n < k. Weconsider the truncated sequence ak−2 < bk−2 < ak−1 < bk−1 < ak < bk . Asremarked above, there should be at least one element of the form cj , which can beinserted between ak−2 and bk , since bk − ak−2 ≥ 6. Let us choose the greatest suchcn, which is cn ≤ bk − 1.
It follows that the set
{a1, a2, . . . , ak, b1, b2, . . . , bk, c1, c2, . . . , cn}is precisely the set of the first consecutive numbers contained between 1 and bk . Inparticular, we have
bk = 2k + n = cn + r, where 1 ≤ r ≤ 5.
On the other hand, we know that cn = 2bn + n − an = bn + n + (bn − an) andbn − an ≤ 2, and so
2k + n = bk = bn + n+ s, where 2 ≤ s ≤ 7.
However, one can improve the upper bound for s as follows. We have equality onlyif r = 5, but in this case we have cn+1 − cn = 6. Using the formula for cn+1 − cn,we derive that bn − an = 1, and thus one has s ≤ 6.
120 6 Algebra and Combinatorics Solutions
Let us now estimate γk , which reads
γk = (1+√3)k−bk = (1+√
3)
(bn + s
2
)− (bn+n+ s) = (s−γn)
(√3 − 1
2
)
.
If s ≤ 6 we use γn ≥ α to obtain
γk ≤ (6 − α)
(√3 − 1
2
)
< β.
If s = 6, then we must have bn+1 −an+1 = an+1 −bn = 2, and therefore bn+1 −bn =4. Thus γn+1 −γn = 1 +√
3 − (bn+1 −bn) = √3 − 3. Since n+ 1 < k, one uses the
induction hypothesis γn+1 > α in order to get γn = γn+1 + 3 − √3 > α + 3 − √
3.Using this inequality above, we derive that
γk ≤ (6 − α − 3 + √3)
(√3 − 1
2
)
= β.
If s = 5, then bn+1 −bn ∈ {3, 4}. Thus γn+1 −γn ≤ √3−2, and so γn > α+2−√
3.Introducing this above, we obtain
γk < (5 − α − 2 + √3)
(√3 − 1
2
)
= β.
The other inequality, γk > α, follows along the same lines.
6.3 Geometric Combinatorics
Problem 2.54. We consider n points in the plane that determine C2n segments, and to
each segment one associates either +1 or −1. A triangle whose vertices are amongthese points will be called negative if the product of numbers associated to its sidesis negative. Show that if n is even, then the number of negative triangles is even.Moreover, for odd n, the number of negative triangles has the same parity as thenumber p of segments labeled −1.
Solution 2.54. Let Pi denote the points in the plane, and let aij be the label of thesegmentPiPj . Then the signature of the trianglePiPjPk is given byPijk = aij ajkaki .In particular, ∏
i �=j �=k �=iPijk = (−1)m,
where m is the number of negative triangles. Now every segment belongs to n − 2triangles, and thus
∏
i �=j �=k �=iPijk =
∏aij ajkaki =
⎛
⎝∏
i �=jaij
⎞
⎠
n−2
= (−1)p(n−2).
Therefore m ≡ pn (mod 2), and the claim follows.
6.3 Geometric Combinatorics 121
Problem 2.55. Given n, find a finite set S consisting of natural numbers larger thann, with the property that, for any k ≥ n, the k × k square can be tiled by a family ofsi × si squares, where si ∈ S.
Solution 2.55. Each of the sets S = {s ∈ Z| n ≤ s ≤ n2}, and S = {s ∈Z|s is prime, n2 < s < 2n2 + n} is convenient, although neither one is minimal.
We will prove this by induction on the size of the square, for the second set S.Consider a k × k square, where k ≥ n2. By hypothesis, the m×m square is tiled bysquares from S, for any m with n ≤ m < k. If k is composite, then write k = pq,where k > p ≥ n. We cover the k × k square by means of q2 p × p squares and weproceed inductively. If k is prime, then k > 2n2 + n. We divide the square into twosquares, onem×m and the other (k−m)× (k−m), and twom× (k−m) rectangles,wherem = n(n+ 1). We have k−m > n2 and thus eachm× (k−m) rectangle canbe covered with m× n or m× (n+ 1) pieces and each of these pieces can be furtherdivided into squares.
Comments 48 The problem whether some region can be tiled by a given set of tilesis difficult and unsolved in general. The interested reader might consult the survey ofArdila and Stanley.
The simpler problem of whether an n×m rectangle was tiled by a×b rectangles wassolved by de Bruijn and, independently, by Klarner in 1969. Specifically, this happensto be possible if and only if mn is divisible by ab, both m and n can be written assums of a’s and b’s, and eitherm or n is divisible by a or elsem (or n) is divisible byb. More generally, them1 ×m2 ×· · ·×mn box is tiled by a1 × a2 ×· · ·× an bricks ifand only if each of the ai has a multiple among themj ’s. Conway and Lagarias gavenecessary conditions for the existence of tilings of rectilinear polygons (i.e., thosehaving sides parallel to the axes) by a set of finite rectilinear tiles using boundaryinvariants, which are combinatorial group-theoretic invariants associated with theboundaries of the tile shapes and the regions to be tiled. Some new invariants werediscovered recently by Pak.
Notice that for n ≥ 2,m > 2, the problem whether there exists a tiling of a givenrectilinear polygon having only horizontal n× 1 tiles and only vertical 1 ×m tiles isan NP -complete question, as shown by Beauquier et al.
• F. Ardila, R. Stanley: Tilings, math.CO/0501170.• D. Beauquier, M. Nivat, E. Rémila, M. Robson: Tiling figures of the plane with
two bars, Comput. Geom. 5 (1995), 1–25.• N. de Bruijn: Filling boxes with bricks, Amer. Math. Monthly 79 (1969), 37–40.• J.H. Conway, J.C. Lagarias: Tiling with polyominoes and combinatorial group
theory, J. Combin. Theory Ser. A 53 (1990), 183–208.• D. Klarner: Packing a rectangle with congruent n-ominoes, J. Combin. Theory 7
(1969), 107–115.• I. Pak: Ribbon tile invariants, Trans.Amer. Math. Soc. 352 (2000), 12, 5525–5561.
Problem 2.56. We consider 3n points A1, . . . , A3n in the plane whose positions aredefined recursively by means of the following rule: first, the triangle A1A2A3 is
122 6 Algebra and Combinatorics Solutions
equilateral; further, the points A3k+1, A3k+2, and A3k+3 are the midpoints of thesides of the triangle A3kA3k−1A3k−2. Let us assume that the 3n points are coloredwith two colors. Show that for n ≥ 7 there exists at least one isosceles trapezoidhaving vertices of the same color.
Solution 2.56. Any subset of three points has two points of the same color. Therefore,there are two points having the same color (which we denote by Aik and Ajk ) among{A3k+1, A3k+2, A3k+3}.
Consider now the lines determined by these monocolor pairs of points:
{Ai1Aj1 , Ai2Aj2 , . . . , AinAjn}.We know that all these directions are parallel to the three sides of the initial equilateraltriangle. Thus there exist at least [n3 ] + 1 ≥ 3 lines in the set above that are parallel.Moreover, each line comes with one of the two colors, and thus there exist at least[
[ n3 ]+12
]+1 ≥ 2 parallel lines of the same color. The four points that determine these
lines form a monocolor isosceles trapezoid.
Problem 2.57. Is there a coloring of all lattice points in the plane using only twocolors such that there are no rectangles with all vertices of the same color, whose sideratio belongs to
{1, 1
2 ,13 ,
23
}?
Solution 2.57. No. According to Van der Waerden’s theorem, there exist four lattice(equidistant) points Ai = (xi, 0) of the same color (say red) on the axis y = 0 suchthat x2 − x1 = x3 − x2 = x4 − x3 = k.
Consider the pointsBi with coordinates (xi, k), lying on a parallel line, at distancek. Then the points Bj contain either two red points—in which case we are done, byconsidering the respective Aj—or else at least three black points, Bj1 , Bj2 , and Bj3 .
Consider next the points Ci = (xi, 2k). Then there exist two points among Cj1 ,Cj2 , and Cj3 that have the same color.
If this color is black, then using the respective Bji and Cji , we obtain a blackrectangle as needed. If the color is red, then the respective Aji and Cji form a redrectangle, as claimed.
Comments 49 Van der Waerden’s theorem from 1927 states that if the set of nat-ural numbers Z+ is partitioned into finitely many color classes, then there existmonochromatic arithmetic progressions of arbitrary length. Erdos and Turán conjec-tured in 1936 the existence of arbitrarily long arithmetic progressions in sequencesof integers with positive density, and their conjecture was successfully confirmed bySzemerédi in 1975. Many extensions have been made since that time. Actually, Sze-merédi proved that for a given length l and density δ > 0, there is an L(l, δ) suchthat if L ≥ L(l, δ), any subsetA ⊂ {1, 2, 3, . . . , L} with more than δL elements, willcontain an arithmetic progression of length l. This makes precise Van der Waerden’stheorem in that there exists W(l, r) such that if W ≥ W(l, r) and {1, 2, 3, . . . ,W }is the union of r subsets, then one of these necessarily contains an l-term arithmeticprogression, and moreover, W(l, r) can be taken as L(l, 1/r).
6.3 Geometric Combinatorics 123
The next breakthrough was made recently by Gowers, who provided the first effectiveupper bound to the function L(l, δ). One consequence is a significant improvementof the upper bound for W(l, r); for example, for some constant c, W(4, r) can betaken as exp(exp(rc)). Generally, we can take L(l, δ) = exp(δ−c(l)), where c(l) =22l+9
. These estimates fall short of providing a proof to another conjecture of Erdosclaiming that any set A ⊂ N with
∑a∈A 1
a= ∞ contains arbitrarily long arithmetic
progressions.
Another far-reaching generalization was provided by Bergelson and Leibman, asfollows. Suppose that p1, p2, . . . , pm are polynomials with integer coefficients andno constant term. Then, whenever Z+ is finitely colored, there exist natural numbersa and d such that the point a and all the points a + pi(d), for 1 ≤ i ≤ m, have thesame color.
Finally, Szemerédi’s theorem was proved to hold for the primes in a spectacular recentpaper by Ben Green and Terence Tao. Specifically, if A is a subset of positive densitywithin the set of primes (for instance the set of all primes), then A contains infinitelymany arithmetic progressions of length k (for any k ≥ 3). An exposition of this resultcan be found in the survey of Kra.
• V. Bergelson, A. Leibman: Polynomial extensions of Van der Waerden’s and Sze-merédi’s theorems, J. Amer. Math. Soc. 9 (1996), 725–753.
• W.T. Gowers: A new proof of Szemerédi’s theorem, Geometric FunctionalAnalysis11 (2001), 465–588.
• B. Kra: The Green–Tao theorem on arithmetic progressions in the primes: anergodic point of view, Bull. Amer. Math. Soc. (N.S.) 43 (2006), 1, 3–23.
• E. Szemerédi: On sets of integers containing no k elements in arithmetic pro-gression, Collection of articles in memory of J.V. Linnik, Acta Arith. 27 (1975),199–245.
• B.L. Van der Waerden: Beweis einer Baudetschen Vermutung, Nieuw Arch. v.Wiskunde 15 (1927), 212–216.
Problem 2.58. Let G be a planar graph and let P be a path in G. We say that Phas a (transversal) self-intersection in the vertex v if the path has a (transversal) self-intersection from the curve-theoretic viewpoint. Let us give an example. Take thepoint 0 in the plane and the segments 01, 02, 03, 04 going counterclockwise around0. Then a path traversing first 103 and then 204 has a (transversal) self-intersection at0, while a path going first along 102 and further on 304 does not have a (transversal)self-intersection.
Prove that any connected planar graphG, with only even-degree vertices, admitsan Eulerian circuit without self-intersections. Recall that an Eulerian circuit is a pathalong the edges of the graph, that passes precisely once along each edge of the graph.
Solution 2.58. We will proceed by double induction, first on the maximum degree ofthe vertices and, for the class of graphs with several vertices of maximum degree, onthe number of such vertices. If the degree is d = 2, then the obvious Eulerian circuitdoes not have self-intersections. Now let v be a vertex of maximal degree d > 2.
124 6 Algebra and Combinatorics Solutions
We split v into two vertices, one of degree 2 and the other of degree d − 2; the firstamong the new vertices is incident to two vertices that were previously incident to v,by means of two consecutive edges (this makes sense since the graph is planar), andthe latter is incident to the remaining ones. If the obtained graph is connected, thenthere exists an Eulerian path without self-intersections, and then we simply restore thegraph and see that this way the circuit remains without self-intersections. If the graphis not connected, then we get two Eulerian circuits. When restoring the graph G, weglue together the two circuits into one Eulerian circuit, still without self-intersections.
Problem 2.59. Let us consider finitely many points in the plane that are not allcollinear. Assume that one associates to each point a number from the set {−1, 0, 1}such that the following property holds: for any line determined by two points fromthe set, the sum of numbers associated to all points lying on that line equals zero.Show that, if the number of points is at least three, then to each point one associatesthe number 0.
Solution 2.59. Let us assume the contrary, i.e., that there exist points with nontrivialnumbers associated to them. Then there should exist points labeled both 1 and −1.
Choose first a point x0 labeled 1. For any other point x �= x0, the line d = x0x
has nε(d) points labeled by ε ∈ {−1, 1}. Moreover, the assumptions imply thatn−1(d) = n+1(d) and n+1(d) ≥ 1, because x0 is labeled 1.
Furthermore, the total number of negative points is n−1 = ∑d n−1(d), the sum
being taken on all lines d passing through x0. Next, the total number of positive pointsis n+1 = 1 + ∑
d(n+1(d) − 1) = n−1 − l + 1, where l is the number of distinctlines passing through x0. We know that l ≥ 2 by hypothesis; hence we obtain thatn+1 ≤ n−1 − 1.
Choose now a point y0 labeled −1. Then the same argument implies that thereversed inequality n−1 ≤ n+1 − 1 holds, contradicting the former inequality.
Problem 2.60. If one has a set of squares with total area smaller than 1, then one canarrange them inside a square of side length
√2, without any overlaps.
Solution 2.60. We put the squares in decreasing order with respect to their sidelengths. The first square is put in the left corner of the square S (the side of which is√
2), the second square will be set to the right of S, and we go on until it becomesimpossible to put another square without surpassing the borders of S. Then, we startanother line of squares above the first square and so on. Assume that the side lengthsare S1 ≥ S2 ≥ · · · and we haven1, n2, . . . , nk squares on the lines number 1, 2, . . . , krespectively. It follows that
√2 − Sn1+1 ≤ S1 + S2 + · · · + Sn1 ≤ √
2,√2 − Sn1+n2+1 ≤ Sn1+1 + · · · + Sn1+n2 ≤ √
2,
and so on. We want to show that
S1 + Sn1+1 + Sn1+n2+1 + · · · ≤ √2,
6.3 Geometric Combinatorics 125
which will prove our claim. We know from above that
S2 + S3 + · · · + Sn1+1 ≥ √2 − S1,
from which we infer
S22 + S2
3 + · · · + S2n1+1 ≥ (√
2 − S1)Sn1+1,
and by the same trick, we obtain
S2n1+2 + · · · + S2
n1+n2+n3≥ (√
2 − S1)Sn1+n2+1,
and so on. By summing up these inequalities, we obtain
1 − S21 ≥ S2
2 + S23 + · · · ≥ (√
2 − S1)(Sn1+1 + Sn1+n2+1 + · · · ),
and hence
S1 + Sn1+1 + Sn1+n2+1 + · · · ≤ 1 − S21√
2 − S1+ S1 = √
2 −(1 − √
2S1)2
√2 − S1
≤ √2.
Comments 50 The problem of packing economically unequal rectangles into a givenrectangle has recently received a lot of consideration. L. Moser asked in 1968 to findthe smallest packing default ε in each of the following situations:
1. the set of rectangles of dimensions 1 × 1n
, for n ∈ Z+, n ≥ 2, whose total areaequals 1 can be packed into the square of area 1 + ε without overlap;
2. the set of squares of side lengths 1n
, for n ∈ Z+, can be packed without overlapinto a rectangle of area π2/6 − 1 + ε;
3. the (infinite) set of squares with sides of lengths 1/(2n + 1), n ∈ Z+, can bepacked in a rectangle of area π2/8 − 1 + ε.
The best results known to this day yield ε very small, at least smaller than 10−9, usinga packing algorithm due to M. Paulhus. Further computational investigations supportthe apparently weaker claim that the packings in the problems above are possible forevery positive number ε. However, G. Martin showed that if this weaker claim holds,then one can also find a packing for ε = 0, in which case the packing is called perfect.The general case of the problem above is still open, but a variation of it has beensolved in the meantime. J. Wästlund proved that if 1/2 < t < 2/3, then the squaresof side n−t , for n ∈ Z+, can be packed into some finite collection of square boxesof the same area ζ(2t) as the total area of the tiles. On the other hand, Chalcraftproved that there is a perfect packing of the squares of side n−3/5 into a square, andpresumably, his technique works for packing the squares of side n−t into a square,where 1/2 < t ≤ 3/5; this is true for packing into a rectangle for all t in the range0.5964 ≤ t ≤ 0.6.
• A. Chalcraft: Perfect square packings, J. Combin. Theory Ser. A 92 (2000), 158–172.
126 6 Algebra and Combinatorics Solutions
• H.T. Croft, K.J. Falconer, R.K. Guy: Unsolved Problems in Geometry, Springer-Verlag, New York, 1994.
• G. Martin: Compactness theorems for geometric packings, J. Combin. TheorySer. A 97 (2002), 225–238.
• A. Meir, L. Moser: On packing of squares and cubes, J. Combin. Theory 5 (1968),126–134.
• M. Paulhus: An algorithm for packing squares, J. Combin. Theory Ser. A 82(1998), 147–157.
• J. Wästlund: Perfect packings of squares using the stack-pack strategy, DiscreteComput. Geom. 29 (2003), 625–631.
Problem 2.61. Prove that for each k there exist k points in the plane, no three collinearand having integral distances from each other. If we have an infinite set of points withintegral distances from each other, then all points are collinear.
Solution 2.61. 1. Consider the pointP1 on the unit circle having complex coordinatesz = 3
5 + 45
√−1. If θ denotes the argument of P1, then θ is noncommensurable withπ , i.e., θ/π ∈ R − Q. Consider the points Pn of coordinates zn. Then the Pn form adense set of points in the unit circle. Moreover, we can compute the distance
|PnPk| = 2| sin(n− k)θ |.Since sin nθ is a polynomial with rational coefficients in sin θ and cos θ , we derivethat the distances between all these points are rational. Given k, choose P1, . . . , Pkand a homothety of large ratio in order to clear the denominators in these distances.We obtain then k points with integral pairwise distances.
2. Given three noncollinear pointsP,Q, andR, with |PQ|, |PR| ≤ k, the numberof points X such that the distances |XP |, |XQ|, |XR| are integral is bounded by4(k + 1)2. In fact, we have, by the triangle inequality,
∣∣|XP | − |XQ|∣∣ < |PQ| = k,
and thus ∣∣|XP | − |XQ|∣∣ ∈ {1, 2, . . . , k}.Moreover, the geometric locus of thoseX for which
∣∣|XP |−|XQ|∣∣ = j is a hyperbolawith foci P and Q. On the other hand,
∣∣|XP | − |XR|∣∣ ∈ {1, 2, . . . , k},and thus X belongs to one of the k + 1 hyperbolas having P and R as foci. ThusX belongs to the intersections of (k + 1)2 pairs of hyperbolas, which have at most4(k + 1)2 intersection points.
Comments 51 This construction of a dense set of points in the unit circle whosepairwise distances are rational is due to A. Muller. Then W. Sierpinski found such adense set in the circle of radius r , provided that r2 is rational, this condition beingnecessary as soon as we have three points at rational distances. The second result isdue to Erdos. The question of Ulam whether there exists a dense set in the plane suchthat the distances between any two of its points is rational is still unsolved.
6.3 Geometric Combinatorics 127
• P. Erdos: Integral distances, Bull. Amer. Math. Soc. 51 (1945), 996.• W. Sierpinski: Sur les ensembles de points aux distances rationnelles situés sur
un cercle, Elem. Math. 14 (1959), 25–27.
Problem 2.62. LetO,A be distinct points in the plane. For each point x in the plane,we write α(x) = xOA (counterclockwise). Let C(x) be the circle of center O andradius |Ox| + α(x)
|Ox| . If the points in the plane are colored with finitely many colors,then there exists a point y with α(y) > 0 such that the color of y also belongs to thecircle C(y).
Solution 2.62. Let G be the graph of vertices {x ∈ E2, α(x) > 0} in which points xand y are adjacent if y ∈ C(x).
If C1(ρ1) and C2(ρ2) are two circles of radii ρ1 < ρ2 < 1 centered at the origin,then there exists a pointM1 ∈ C1(ρ1) that is adjacent to all points ofC2(ρ2). The pointM1 has α(M1) = θ uniquely determined, as follows. The conditionC(M1) = C2(ρ2)
is equivalent to ρ2 = ρ1 + θρ1
, yielding
θ = (ρ2 − ρ1)ρ1 ∈ (0, 1) ⊂ (0, 2π),
and therefore M1 is well determined.Assume now that the claim from the statement is not true. It will follow that the
color of M1 is distinct from the colors appearing on the circle C2(ρ2).Let now C1, C2, . . . , Ck, . . . be an infinite sequence of circles centered at the
origin whose respective radii are
0 < ρ1 < ρ2 < · · · < ρk < · · · < 1.
Then let the set of colors that we encounter on Ck be denoted by τk. The previousargument, when applied to the pair of circles Ci and Cj , shows that there exists apoint on Ci that is colored by a color that does not exist on Cj , if i < j , and thusτi \τj �= ∅. Since the number of colors is finite, this is impossible as soon as k is largeenough (e.g., k > 2p, where p is the number of colors). This contradiction provesthe claim.
Problem 2.63. Let k, n ∈ Z+.
1. Assume thatn−1 ≤ k ≤ n(n−1)2 . Show that there existn distinct points x1, . . . , xn
on a line, that determine exactly k distinct distances |xi − xj |.2. Suppose that
[n2
] ≤ k ≤ n(n−1)2 . Then there exist n points in the plane that
determine exactly k distinct distances.3. Prove that for any ε > 0, there exists some constant n0 = n0(ε) such that for anyn > n0 and εn < k <
n(n−1)2 , there exist n points in the plane that determine
exactly k distinct distances.
Solution 2.63. 1. Let m be an integer between 1 and n− 1 such that
n− 1 = C2n − C2
n−1 ≤ k ≤ C2n − C2
m.
128 6 Algebra and Combinatorics Solutions
Set p = k− (m− 1)−C2n +C2
m+1; in particular, we have 1 ≤ p ≤ m. Consider thenthe following set of points of the real line:
X = {1, 2, . . . , m,m+ p} ∪ {πm+2, πm+3, . . . , πn
}.
The distances between the firstm+1 points correspond to the set {1, 2, . . . , p+m−1}.The remaining points are independent transcendental points, and thus the distances be-tween them are all distinct transcendental numbers. Moreover, the distances betweenpoints of the second set and points of the first set are also all distinct (translates ofthe former set of transcendental distances by integers). Thus the number of distancesobtained so far is
p +m− 1 +(
n∑
i=m+2
(i − 1)
)
= k.
2. If[n2
] ≤ k ≤ C2n , we put the n points in the consecutive vertices of a regular
(2k+ 1)-gon. There are exactly k distinct distances in a regular (2k+ 1)-gon, and allof them are realized at least once, since k ≥ [
n2
].
3. It suffices to check the case εn ≤ k ≤ [n2
]. Let m = 3 + [√
n]. For m ≤ a ≤
n− 2m, 1 ≤ b ≤ m, we consider the following subset of Z × Z:
X = C ∪ {(i, 1)|1 ≤ i ≤ a − 1} ∪ {(1, j)|2 ≤ j < m} ∪ {(0, h)|1 ≤ h ≤ b},where C is a subset of {(1, j), 2 ≤ i ≤ a − 1, 2 ≤ j ≤ m} with the property that thedistances between the pairs of points in C are all distinct.
The set of squares of the distances between points in X is
D(a, b) = {(i2 + j2)|0 ≤ i ≤ a − 1, 0 ≤ j ≤ m− 1} ∪ {a2 + j2|0 ≤ j ≤ b − 1}.We have
D(m, 1) ⊂ D(m, 2) ⊂ · · · ⊂ D(m,m) ⊂ D(m+ 1, 1)
⊂ D(m+ 1, 2) ⊂ · · · ⊂ D(n− 2m,m),
and each set from above is obtained from the previous one in the sequence by adjoiningat most one more distance; thus the difference between consecutive sets is either emptyor a singleton. Therefore, card(D(a, b)) is an increasing sequence of consecutiveelements that lie between card(D(m, 1)) and card(D(n− 2m,m)). Observe now thatthe upper bound card(D(n− 2m,m)) is at least n− 2m− 1 ≥ [
n2
].
Further, the set D(m, 1) consists of integers between 1 and 2m2 (note that we havethe asymptotic behavior m2 ∼ n) that are sums of two perfect squares. It is knownthat whenever μ = λ2 + δ2 and p is a prime number p ≡ 3 (mod 4) if p divides μ,then p2 also divides μ. Thus, the density of the set of numbers that are sums of twoperfect squares is bounded by the product over the primes p
∏
p≡3 (mod 4)
(1 − 1
p+ 1
p2
),
which tends to 0. Therefore, for n large enough, we have card(D(m, 1)) < εn, whichproves the claim.
6.3 Geometric Combinatorics 129
Comments 52 The result is due to P. Erdos, who proposed it in 1980 as a problemin Amer. Math. Monthly, related to another question that he considered long ago. In1946, Erdos posed the problem of determining the minimum number d(n) of differentdistances determined by a set of n points in R
2, proved that d(n) ≥ cn1/2, andconjectured that d(n) ≥ cn/
√log n. If true, this inequality is best possible, as shown
by the lattice points of the plane. The best result known is due to J. Solymosi andCs.D. Tóth, and yields d(n) > cn6/7.
• P. Erdos: On the set of distances of n points, Amer. Math. Monthly 53 (1946),248–250.
• P. Erdos: Problem 6323, Amer. Math.Monthly 87 (1980), 826.• J. Solymosi, Cs.D. Tóth: Distinct distances in the plane, The Micha Sharir birthday
issue, Discrete Comput. Geom. 25 (2001), 629–634.
Problem 2.64. Show that it is possible to pack 2n(2n + 1) nonoverlapping pieceshaving the form of a parallelepiped of dimensions 1 × 2 × (n+ 1) in a cubic box ofside 2n+ 1 if and only if n is even or n = 1.
Solution 2.64. Let us consider the cube composed of (2n + 1)3 cells of dimensions1×1×1, which are labeled using the coordinate system (a1, a2, a3), ai ∈ {1, . . . , 2n+1}.
1. Case n = 1. The 6 pieces are arranged as follows. The first three pieces aregiven by
x1 = {(1, 1, 2), (1, 1, 3), (1, 2, 2), (1, 2, 3)},x2 = {(1, 2, 1), (1, 3, 1), (2, 2, 1), (2, 3, 1)},x3 = {(2, 1, 1), (3, 1, 1), (2, 1, 2), (3, 1, 2)}.
The next three pieces x4, x5, x6 are located as above, by permuting the indices 3 and1. The cells (1, 1, 1), (2, 2, 2), (3, 3, 3) remain empty.
2. Suppose now that n is even. We have then (2n+1)3−2n(2n+1)·1·2·(n+1) =2n+1 cases that remain empty. Each sliceP1,i = {(i, a, b)|a, b ∈ {1, 2, . . . , 2n+1}}can be divided in four smaller boxes P1,i;1 = {(i, a, b)|a ≤ n+ 1, b ≤ n}, P1,i;2 ={(i, a, b)|n+ 2 ≤ a, b ≤ n+ 1}, P1,i;3 = {(i, a, b)|n+ 1 ≤ a, n+ 2 ≤ b}, P1,i;4 ={(i, a, b)|a ≤ n, n+ 1 ≤ b} and one extra cell, namely {(i, n+ 1, n+ 1)}.
Each box P1,i;s has dimensions 1 × n× (n+ 1) and can be covered by n2 pieces
1 × 2 × (n+ 1). The remaining cells of coordinates (i, n+ 1, n+ 1) remain empty.3. Assume further that n is odd, n ≥ 3, and that there exists a packing as in
the statement. One associates to a cell (a1, a2, a3) the number p, which counts thenumber of i ∈ {0, 1, 2, 3} such that ai = n + 1. One colors the cell blue if p = 3,red if p = 2, yellow when p = 1, and leaves it colorless when p = 0. Then, we haveone blue cell and 6n red cells.
Now, for each i ≤ 2n+ 1, k ∈ {1, 2, 3}, we set Pk,i = {a1, a2, a3), ak = i}.Then each slice Pk,i contains an odd number of cells (2n+ 1)2. In the meantime,
every piece contains an even number of cells from a given slice Pk,i , which might be
130 6 Algebra and Combinatorics Solutions
2, n+ 1, or 2(n+ 1) cells, respectively. This means that in each slice Pk,i there existsat least one empty cell.
The total number of empty cells is 2n+1, and for fixed k the slicesPk,1, . . . , Pk,2n+1are disjoint, so that each slice Pk,i contains exactly one empty cell. In particular, theset consisting of the blue cell and the red ones contains at most one empty cell.
Furthermore, each piece S contains colored cells because n+1 > 12 (2n+1), and
it fits into one of the cases below:Type 1. S contains the blue cell, n+ 1 red cells, and n yellow cells.Type 2. S does not contain the blue cell, and it contains 2 red and 2n yellow cells.Type 3. S does not contain the blue cell, and it contains 1 red, n+ 1 yellow, and
n colorless cells.Type 4. S does not contain either the blue cell or red cells, but 2 yellow and 2n
colorless cells.Let us analyze now the total number of colored cells which can be covered by all thepieces.
• The blue cell is empty. Then there are nonempty red cells and we compute thatthe 2n(2n+ 1) pieces contain 2n(2n+ 1) · 2 + 6n · n = 14n2 + 4n colored cells.
• One red cell is empty. Then there exists a piece S of type 1 that contains n+ 1 redcells and 4n2 + 2n− 1 pieces of type 2. Altogether they contain 5n− 2 red cells.Then the pieces contain 2n+ 2 + (4n2 + 2n− 1) · 2 + (5n− 2)n = 13n2 + 4ncolored cells.
• The blue and red cells are not empty. Then the pieces of soap will contain nmorecolored cells than in the previous case.
However, the total number of colored cells within the cube is 1 + 6n+ 12n2, and forn ≥ 3, we have the inequality 13n2 + 4n > 12n2 + 6n+ 1, which is a contradiction,because the pieces are supposed to be nonoverlapping.
4. Second proof when n > 3. For k ∈ {1, 2, 3}, let us denote bymk the number ofpieces whose longest side is orthogonal to the plane of the slice Pk,i . Each such piecehas exactly 2 cells in common with Pk,n+1. Every piece from the remaining ones haseither n+ 1 or 2(n+ 1) cells in common with Pk,n+1. Therefore, n+ 1 should dividethe number of remaining cells, which is 4n(n+ 1)− 2mk = (2n+ 1)2 − 1 − 2mk .This implies that n+ 1 divides mk for all k.
Moreover, the total number of cells is m1 + m2 + m3 = 2n(2n + 1). Sincegcd(n, n+ 1) = gcd(n, 2n+ 1) = 1, we derive that n ∈ {1, 3}.Problem 2.65. Let F be a finite subset of R with the property that any value of thedistance between two points from F (except for the largest one) is attained at leasttwice, i.e., for two distinct pairs of points. Prove that the ratio of any two distancesbetween points of F is a rational number.
Solution 2.65. Let F = {s1, . . . , sn} be a finite set from a vector space V over Q.Consider the n(n − 1)/2 difference vectors si − sj , where i < j . Some differencevectors appear several times in the difference vectors sequence, and we call them Dvectors. We will prove a more general result: The vector space generated by those
6.3 Geometric Combinatorics 131
differences that are not D vectors coincides with the vector space generated by alldifference vectors.
Proof of the claim. Let us assume the contrary. By the Hahn–Banach lemma, thereexists a linear functional f : V → Q that vanishes on those vectors that are not Dvectors, although f is nonzero on all difference vectors. Observe that we can replaceV with a finite-dimensional linear subspace that contains F .
Let M = f (si),m = f (sj ) be the greatest and respectively the smallest valuesfrom f (F ), where of course, m �= M . The set of linear functionals f : V → Q thatmap F one-to-one into Q is dense in the set of linear functionals. Consequently, thereexists g : V → Q that injects F into Q such that
|f (s)− g(s)| ≤ 1
5(M −m), for all s ∈ F.
Consider g(sp) the maximal value and g(sq) the minimal value within the set g(F ) ⊂Q. Therefore sp − sq is not a D vector and thus f vanishes: f (sp − sq) = 0. Thisimplies that
g(si)− g(sj ) ≤ g(sp)− g(sq) ≤ f (sp)− f (sq)+ 2
5(M −m) = 2
5(M −m).
On the other hand, g is closely approximated by f ; hence
g(si)− g(sj ) ≤ g(sp)− g(sq) ≤ f (sp)− f (sq)+ 2
5M − n = 2
5(M − n),
and also
g(si)− g(sj ) ≥ f (si)− 1
5(M −m)−
(f (sj )+ 1
5(M −m)
)= 3
5(M −m),
contradicting the previous equality. This proves our general claim.Assume now that F ⊂ R, and the extreme points of F are 0 and 1. Consider then
V = R as a vector space over Q. From the hypothesis, 1 is the only distance that isnot attained twice. The claim above tells us that the Q-linear space generated by 1contains all difference vectors. Thus all distances are rational.
Comments 53 The Hahn–Banach lemma, alluded to above, says that for any vectorsubspace W of the vector space V and any vector v ∈ V that does not belong to W ,there exists a linear map f : V → R such that f (W) = 0 but f (v) �= 0.
The result from the statement is due to Mikusinski and Schinzel, and the proof givenabove is due to Straus.
• J. Mikusinski, A. Schinzel: Sur la réductibilité de certains trinômes, Acta Arith.9 (1964), 91–95.
• E.G. Straus: Rational dependence in finite sets of numbers, Acta Arith. 11 (1965),203–204.
7
Geometry Solutions
7.1 Synthetic Geometry
Problem 3.1. Let I be the center of the circle inscribed in the triangle ABC andconsider the points α, β, γ situated on the perpendiculars from I on the sides of thetriangle ABC such that
|Iα| = |Iβ| = |Iγ |.Prove that the lines Aα,Bβ,Cγ are concurrent.
Solution 3.1. The lines Aα,Bβ,Cγ are concurrent if and only if they satisfy Ceva’stheorem. Let us draw A1A2 ‖ BC, where A1 ∈ |AB|, A2 ∈ |AC|, and α ∈ |A1A2|.We also draw their analogues. Using Thales’ theorem, the Ceva condition is reducedto ∏
cyclic
|αA2||αA1| = 1.
Now I is situated on all three bisectors and |Iγ | = |Iα|. By symmetry we have|C1γ | = |A2α| and their analogues. This shows that the Ceva condition is satisfied.
Problem 3.2. We consider the angle xOy and a pointA ∈ Ox. Let (C) be an arbitrarycircle that is tangent to Ox and Oy at the points H and D, respectively. Set AE forthe tangent line drawn from A to the circle (C) that is different from AH . Show thatthe lineDE passes through a fixed point that is independent of the circle (C) chosenabove.
Solution 3.2. Let A′ be the point on Oy such that |OA′| = |OA|. Let us furtherconsider the intersection points AE ∩Oy = {F } and DE ∩ |AA′| = {P }.
134 7 Geometry Solutions
O
A
A'
DF
EP
H
We use Menelaus’s theorem in the triangle AFA′ with the line DEP as transversal.We have |PA′|
|PA| · |EA||EF | · |DF |
|DA′| = 1.
But one knows that |EA| = |AH |, |EF | = |DF |, and |A′D| = |AH |. Then |PA′||PA| =
1, and therefore P is the midpoint of |AA′|. Consequently, P is a fixed point.
Problem 3.3. Let C be a circle of center O and A a fixed point in the plane. For anypoint P ∈ C, let M denote the intersection of the bisector of the angle AOP withthe circle circumscribed about the triangle AOP . Find the geometric locus of M asP runs over the circle C.
Solution 3.3. One knows that OM is the bisector of AOP , and therefore OM is themediator of the segment PB. Also,M is the midpoint of the arc of the circle AP , andthereforeM belongs to the mediator of the segment |AP |. In the triangle PAB,M isthe intersection of two mediators, and hence it also belongs to the mediator of segment|AB|. It is immediate now that the geometric locus of M is the entire mediator.
Problem 3.4. Let ABC be an isosceles triangle having |AB| = |AC|. If AS is aninterior Cevian that intersects the circle circumscribed aboutABC at S, then describethe geometric locus of the center of the circle circumscribed about the triangle BST ,where {T } = AS ∩ BC.
Solution 3.4. We have BST = BCA = CBA. Thus the line AB is tangent to thecircle circumscribed about BST . Consequently, the center of this circle is situated onthe line that passes through B that is perpendicular to AB.
7.1 Synthetic Geometry 135
Problem 3.5. LetAB,CD,EF be three chords of length one on the unit circle. Thenthe midpoints of the segments |BC|, |DE|, and |AF | form an equilateral triangle.
Solution 3.5. Leta, b, c, d, e, f be the complex affixes (coordinates) of the respectivepoints on the unit circle and ε = exp
( 2πi3
). Then the triangles OAB,OCD,OEF
are equilateral, and therefore we have the following complex identities:
a + εb = 0,
εc + ε2d = 0,
ε2e + f = 0.
These imply that a+f2 + ε b+c2 + ε2 d+c2 = 0. Now the midpoint of |AF | has the affix
a+f2 , and the other midpoints are similarly associated to b+c
2 and d+e2 . The relation
above is equivalent to the claim of the statement.
Problem 3.6. Denote by P the set of points of the plane. Let � : P × P → P bethe following binary operation: A � B = C, where C is the unique point in the planesuch that ABC is an oriented equilateral triangle whose orientation is counterclock-wise. Show that � is a nonassociative and noncommutative operation satisfying thefollowing “medial property”:
(A � B) � (C � D) = (A � C) � (B � D).
Solution 3.6. Let A �= B,A � B = C. Then (A � B) � C = C � C = C. ButA � (B � C) = A �A = A �= C; therefore the operation � is nonassociative. BecauseA � B,B � A are symmetric with respect to AB, the operation � is noncommutative.
Now identify the points of the plane with complex numbers. Set ε = exp( 2πi
3
).
We have then A � B = A+ ε(B − A). It is now easy to verify that
(A � B) � (C � D) = (1 − ε)2A+ ε(1 − ε)(B + C)+ ε2D.
Moreover, this expression is symmetric in B and C.
Comments 54 It is possible to take an arbitrary ε ∈ C and to define A �ε B = C,where C is the point such that the triangle ABC is similar to the triangle determinedby the points 0, 1, and ε. Again, the medial property holds for this binary law.
Problem 3.7. Consider two distinct circles C1 and C2 with nonempty intersectionand let A be a point of intersection. Let P,R ∈ C1 and Q,S ∈ C2 be such thatPQ and RS are the two common tangents. Let U and V denote the midpoints of thechords PR and QS. Prove that the triangle AUV is isosceles.
Solution 3.7. Let B be the other intersection point of the two circles. It is knownthat the line AB is the radical axis of the two circles, i.e., the geometric locus of thepoints having the same power with respect to the two circles. Let us then consider theintersection point AB ∩ |PQ| = X. The power of X with respect to either circle isPC1(X) = |XA| · |XB|. On the other hand, PC1(X) = |XP |2 and PC2(X) = |XQ|2,
136 7 Geometry Solutions
and so we obtain that X is the midpoint of |PQ|. Thus the line AB passes throughthe midpoints of |PQ| and |RS|, and thus it is also passes through the midpoint of|UV |, being also orthogonal to the latter. This shows that AUV is isosceles.
Problem 3.8. If the planar triangles AUV, VBU , and UVC are directly similar to agiven triangle, then so is ABC. Recall that two triangles are directly similar if onecan obtain one from the other using a homothety with positive ratio, rotations andtranslations.
Solution 3.8. Using complex numbers, each similarity (i.e., homothety) has the formz �→ αiz + βi , for some αi, βi ∈ C. The fixed triangle has its vertices located atz1, z2, and z3, and the pointsA,B,C,U, V are located at a, b, c, u, v respectively. Byhypothesis, u = αizi+1 + βi and v = αizi+2 + βi , where the indices are consideredmod 3. This implies that αizi(zi+1 − zi+2) = zi(u − v) and βi(zi+1 − zi+2) =zi+1v − zi+2u, so that
3∑
i=1
(αizi + βi)(zi+1 − zi+2) = 0.
But from3∑
i=1
(zi+1 − zi+2) = 0,3∑
i=1
zi(zi+1 − zi+2) = 0
we derive
det
⎛
⎝a z1 1b z2 1c z3 1
⎞
⎠ = det
⎛
⎝α1z1 + β1 z1 1α2z2 + β2 z2 1α3z3 + β3 z3 1
⎞
⎠ = 0,
which is equivalent to saying that ABC is similar to the fixed triangle.
Problem 3.9. Find, using a straightedge and a compass, the directrix and the focusof a parabola. Recall that the parabola is the geometric locus of those points P in theplane that are at equal distance from a point O (called the focus) and a line d calledthe directrix.
Solution 3.9. Let� be an arbitrary direction in the plane. We draw two distinct lines�1 and �2 cutting the parabola in two nontrivial chords X1X2 and Y1Y2, both beingparallel to �. Let X, Y be the midpoints of |X1X2| and respectively |Y1Y2|. It issimple to show that the line XY (called the diameter conjugate to the direction �)is parallel to the symmetry axis of the parabola. Moreover, let T� be the intersectionpoint between the lineXY and the parabola. Then the line that passes through T� andis parallel to � is tangent (at T�) to the parabola.
Now we will use a well-known theorem in the geometry of conics, which claimsthat if P is a point in the plane (outside the convex hull of the parabola), PA, PBare the two tangents to the parabola issued from P (where A and B belong to theparabola) and if the angle APB = π
2 , then AB is a focal chord, i.e., the focus F lieson the chord |AB|.
7.1 Synthetic Geometry 137
Choose now the direction �⊥ that is orthogonal to � and construct as above thepoint T�⊥ on the parabola. Then, according to the previous result, the chord T�T�⊥is focal.
Choose next another couple of orthogonal directions, δ and δ⊥, and derive the newfocal chord TδTδ⊥ . Then the two focal chords intersect at the focus T�T�⊥ ∩TδTδ⊥ ={O}. Moreover, we can draw the axis of symmetry of the parabola as the line passingthrough the focus that is parallel to (any) diameter conjugate to XY .
Finally, let S be the intersection of the axis with the parabola and let O ′ be thesymmetric of the focus with respect to S. Then the directrix d is the line passingthrough O ′ that is orthogonal to the axis OSO ′.
Then �, the conjugate of �, meets � at a point on the axis. We represent then�, � and we find two points of the axis.
Problem 3.10. Prove that if M is a point in the interior of a circle and AB ⊥ CD
are two chords perpendicular at M , then it is possible to construct an inscribablequadrilateral with the following lengths:
∣∣|AM| − |MB|∣∣, |AM| + |MB|, ∣∣|DM| − |MC|∣∣, |DM| + |MC|.Solution 3.10. Let us construct AF ‖ CD and CE ‖ AB, where the points E,F lieon the given circle.
138 7 Geometry Solutions
Then the intersection point BF ∩ ED = O is the center of the circle. Moreover, wecan compute |CE| = ∣∣|MB| − |MA|∣∣ and |AF | = ∣∣|MD| − |MC|∣∣. We rotate thetriangle CED around the center O in order to superpose ED onto FB. Call C′ thenew position of the point C. Then, the length of the fourth side of the quadrilateralABC′F is |BC′| = |DM| + |MC|.Problem 3.11. If the Euler line of a triangle passes through the Fermat point, thenthe triangle is isosceles.
Solution 3.11. The Euler lineOG is determined by the circumcenterO and the cen-ter of gravity G. The trilinear coordinates of the points involved are G(sinB sinC,sinA sinC, sinA sinB), O(cosA, cosB, cosC), and the Fermat point F(sin(B +60), sin(C + 60), sin(A + 60)). Therefore, G,O, and F are collinear only if thedeterminant of their trilinear coordinates vanishes. By direct calculation, this deter-minant can be computed as
sin(A− B) sin(B − C) sin(C − A) = 0,
and hence the triangle must be isosceles.
Problem 3.12. Consider a point M in the interior of the triangle ABC, and chooseA′ ∈ AM,B ′ ∈ BM , and C′ ∈ CM . Let P,Q,R, S, T , and U be the intersectionsof the sides of ABC and A′B ′C′. Show that PS, TQ, and RU meet at M .
Solution 3.12. The triangles ABC and A′B ′C′ are in perspective with centerM , andhence A′C′ ∩ AC,B ′C′ ∩ BC, and B ′A′ ∩ BA are collinear. Let xy be the line onwhich these points lie.
We make a projective transformation that sends the line xy into the line at infinity.This way, the sides of the two triangles become pairwise parallel.
Consider A′ fixed and let the points B ′′, C′′ be mobile points sitting on AB ′ andAC′, respectively. If we move the line B"C" by keeping it parallel to BC, then theintersection pointsQ′ = AB ∩B ′′C′′ and P ′ = AC ∩B ′′C′′ determine homographicdivisions on AB and AC. When we reach the position where A ∈ B ′′C′′, we see thatQ′T ∩ SP ′ = A; also, when we reach the position for which A′ ∈ B ′′C′′, we find
7.1 Synthetic Geometry 139
thatQ′T ∩ SP ′ = A′. Therefore, sinceQ′, P ′ determine homographic divisions, weobtain that for any position of B ′′C′′, we have Q′T ∩ SP ′ ∈ AA′. In particular, thishappens when B ′′C′′ = B ′C′, and thus QT ∩ SP ∈ AA′. Analogously, we haveQT ∩ RU ∈ CC′, SP ∩ RU ∈ BB ′. According to Pascal’s theorem, the diagonalsof the hexagon PQRSTU are concurrent, since the edges are pairwise parallel. Thisshows that QT ∩ SP ∩ RU ∈ AA′ ∩ BB ′ ∩ CC′ = M .
Problem 3.13. Show that if an altitude in a tetrahedron crosses two other altitudes,then all four altitudes are concurrent.
Solution 3.13. If the altitude hA fromA intersects hB , thenAB⊥CD. Similarly, if hA
intersects hC , thenAC⊥BD. These can be written vectorially as 〈 →AB,
→AD − →
AC〉 =0, and 〈 →
AC,→AD − →
AB〉 = 0, where 〈, 〉 denotes the Euclidean scalar product. These
imply that 〈 →AB − →
AC,→AD〉 = 0, which is equivalent to AD⊥BC. This implies that
all four altitudes are concurrent.
Problem 3.14. Three concurrent Cevians in the interior of the triangle ABC meetthe corresponding opposite sides atA1, B1, C1. Show that their common intersectionpoint is uniquely determined if |BA1|, |CB1|, and |AC1| are equal.
Solution 3.14. Let |BA1| = |CB1| = |AC1| = δ. From Ceva’s theorem, we obtain
(a − δ)(b − δ)(c − δ) = δ3.
We may assume that a ≤ b ≤ c, and hence δ ∈ [0, a].The left-hand side is a decreasing function of δ ranging from the value abc,
obtained for for δ = 0, to 0, when δ = a. Further, the right-hand-side function δ3 isstrictly increasing. Therefore, the two functions can have only one common value onthe interval [0, a]. This determines uniquely the intersection point.
Comments 55 Also, if we ask that |CA1|, |BC1|, and |AB1| be equal, their commonvalue δ coincides with the δ we found above. In particular, the two intersection pointsare isotomic to each other.
Problem 3.15. LetABCD be a convex quadrilateral with the property that the circleof diameter AB is tangent to the line CD. Prove that the circle of diameter CD istangent to the line AB if and only if AD is parallel to BC.
Solution 3.15. Let M be the midpoint of |AB| and draw the perpendicular MM ′ ⊥CD, where M ′ ∈ CD. Then M ′ is the contact point of the circle C1 of diameter ABand the line CD. Thus AM ′B = π
2 , and so |MM ′| = |AB|2 . Conversely, ifM,M ′ are
as above and satisfy the condition |MM ′| = |AB|2 , then the circle of diameter AB is
tangent to CD.
140 7 Geometry Solutions
Let nowO = AB∩CD, assuming that the two lines are not parallel. If |OM| = λ
and AOD = θ , then |MM ′| = λ sin θ and so |MB| = |MA| = λ sin θ . This turns to
|OB||OA| = |OM| + |MA|
|OM| − |MB| = λ(1 + sin θ)
λ(1 − sin θ)= 1 + sin θ
1 − sin θ.
Conversely, |OB||OA| = 1+sin θ
1−sin θ implies that |MM ′| = |MA| = |MB|.Let now P be the midpoint of CD and draw the perpendicular PP ′ ⊥ AB,
with P ′ ∈ AB. The previous argument shows that |PP ′| = |CD|2 is equivalent to
|OC||OD| = 1−sin θ
1+sin θ , which, according to Thales’ theorem, is equivalent to saying thatAD
is parallel to BC. But |PP ′| = |CD|2 iff the circle of diameter CD is tangent to AB,
as claimed.The last case occurs when AB is parallel to CD. Then |MM ′| is the distance
between the lines AB and CD, and hence |MM ′| = |AB|2 = |CD|
2 . Thus ABCD is aparallelogram, and hence AD is parallel to BC.
Problem 3.16. Let A′, B ′, C′ be points on the sides BC,CA,AB of the triangleABC. LetM1,M2 be the intersections of the circleA′B ′C′ with the circleABA′ andlet N1, N2 be the analogous intersections of the circle A′B ′C′ with the circle ABB ′.
1. Prove thatM1M2, N1N2, AB are either parallel or concurrent, in a point that wedenote by A′
1;2. Prove that the analogously defined points A′
1, B′1, C
′1 are collinear.
Solution 3.16. 1. More precisely,M1M2, N1N2, AB are the radical axes of the threepairs of circles formed by A′B ′C′, ABB ′, and ABC′. The first part follows.
2. Further, A′1 is the radical center of A′B ′C′, ABB ′, ABA′, and thus |A′
1A| ·|A′
1B| = pA′
1ABB ′ = p
A′1
A′B ′B ′ , where p denotes the power of the point with respect to
the given circle. Therefore pA′
1A′B ′C′ = p
A′1
ABC , and their analogues, which imply thatthe three pointsA′
1, B′1, andC′
1 belong to the radical axis of the pair of circlesA′B ′C′and ABC.
Problem 3.17. A circumscribable quadrilateral of area S = √abcd is inscribable.
Solution 3.17. By the circumscribability hypothesis, we have a + c = b + d . Let kbe the length of the diagonal that leaves a, b on one side and c, d on the other side.
7.1 Synthetic Geometry 141
Let α, β be the angles of the quadrilateral between a and b, and c and d, respectively.We have then the law of cosines computing k:
k2 = a2 + b2 − 2ab cosα,
k2 = c2 + d2 − 2cd cosβ.
Subtracting the terms (a − b)2 = (c − d)2, we obtain
2ab(1 − cosα) = 2cd(1 − cosβ).
Now compute the area S of the quadrilateral S = 12 (ab sin α+ cd sin β), which is by
hypothesis√abcd. Therefore
4S2 = 4abcd = a2b2(1 − cos2 α)+ c2d2(1 − cos2 β)+ 2abcd sin α sin β.
From the former identity, we obtain
4abcd = ab(1+cosα)cd(1−cosβ)+cd(1+cosβ)ab(1−cosα)+2abcd sin α sin β.
After simplification of the terms, we obtain
4 = 2 − 2 cos(α + β),
and therefore α + β = π . This implies that the quadrilateral is inscribable.
Problem 3.18. Let O be the center of the circumcircle, Ge the Gergonne point, Nathe Nagel point, andG1, N1, the isogonal conjugates ofG andN , respectively. ProvethatG1, N1, andO are collinear (see also the Glossary for definitions of the importantpoints in a triangle).
Solution 3.18. We will use the trilinear coordinates with respect to the given trian-gle. These are triples of numbers (see the triangle compendium from the Glossary)proportional to the distances from the given point to the edges of the triangle. Thetrilinear coordinates behave very nicely when we consider the isogonal conjugate. If
P has the coordinates (x, y, z), then its isogonal conjugate has coordinates(
1x, 1y, 1z
).
The trilinear coordinates of the points considered above are then
O(cosA, cosB, cosC),
Ge
(1
a(p − a),
1
b(p − b),
1
c(p − c)
), G1 (a(p − a), b(p − b), c(p − c)) ,
Na
(p − a
a,p − b
b,p − c
c
), N1
(a
p − a,
b
p − b,
c
p − c
).
Further, three points are collinear if the determinant of the 3 × 3 matrix of theircoordinates vanishes. The claim from the statement is then equivalent to the followingidentity:
142 7 Geometry Solutions
det
⎛
⎝cosA cosB cosC
a(p − a) b(p − b) c(p − c)a
p−ab
p−bc
p−c
⎞
⎠ = 0.
We develop the determinant and obtain
∑
cyclic
cosA bc
(p − b
p − c− p − c
p − b
)=
∑
cyclic
cosAbc
(p − b)(p − c)a(c − b).
This vanishes iff ∑
cyclic
cosA (p − a)(c − b) = 0,
which is equivalent to
∑
cyclic
(b2 + c2 − a2)(c − b)(−a + b + c)a = 0,
which is immediate.Note that we could use instead the barycentric coordinates as well. These coordi-
nates are triples of numbers proportional to the areas of the three triangles determinedby the point and two of the vertices of the initial triangle. The barycentric coordinatesof the points considered above are given by
O(a, b, c), Ge
(1
p − a,
1
p − b,
1
p − c
), G1
(a2(p − a), b2(p − b), c2(p − c)
),
Na (p − a, p − b, p − c) , N1
(a2
p − a,b2
p − b,c2
p − c
).
Finally, three points are collinear if and only if their coordinate matrix has vanishingdeterminant.
7.2 Combinatorial Geometry
Problem 3.19. Consider a rectangular sheet of paper. Prove that given any ε > 0,one can use finitely many foldings of the paper along its sides in either 2 equal partsor 3 equal parts to obtain a rectangle whose sides are in ratio r for some r satisfying1 − ε ≤ r ≤ 1 + ε.
Solution 3.19. Observe that log3 2 �∈ Q and thus the residues modulo 1 of the ele-ments n log3 2, with n ∈ Z+, form a dense subset of (0, 1). This implies that the setof numbers of the form 2a3b, with a, b ∈ Z, are dense in R.
Problem 3.20. Show that there exist at most three points on the unit disk with thedistance between any two being greater than
√2.
7.2 Combinatorial Geometry 143
Solution 3.20. Consider the pointsAi on the unit disk of radius 1 and centerO. Then|OAi | ≤ 1, and if we assume that |AiAj | >
√2, then 2 < |OAi |2 + |OAj |2 −
2|OAi | · |OAj | cos AiOAj . Thus, cos AiOAj < 0 and hence AiOAj > π2 , which
implies that n ≤ 3.
Comments 56 H.S.M. Coxeter proposed this problem in 1933. The same argumentshows that if we have k pointsAi within the unit disk, then there exists a pair of pointssuch that |AiAj | ≤ max
(1, 2 sin π
k
), and this is sharp for 2 ≤ k ≤ 7.
The result was generalized to higher dimensions by Davenport, Hajós, and, indepen-dently, Rankin, as follows: There are at most n+ 1 points in the unit ball of R
n suchthat the distance between any two points is greater than
√2.
Here is a proof sketch. Let Ai denote the m points satisfying the hypothesis, so thatthey are different from the origin O. Then we have m ≥ n + 2 unit vectors OAi|OAi |in R
n with the property that 〈vi, vj 〉 < 0. We claim next that there exist pairwiseorthogonal subspaces W1,W2, . . . ,Wm−n ⊂ R
n that span Rn such that each space
Wj contains 1+dimWj vectors among the unit vectors above, v1, . . . , vm, that spanWj . Since m − n ≥ 2, there exist two vectors among the vi that are orthogonal toeach other, which contradicts our assumptions. The claim follows by induction on thedimension n. If n = 2, it is immediate. Now, if vi = −vn, for some i < n, then allother vj should be orthogonal to vn (and to vi). Otherwise, we can write for all i < n,vi = ui + xivn, for some scalars xi < 0 and some nonzero vector ui orthogonal tovn. This implies that 〈ui, uj 〉 < 0, when i �= j . We can apply therefore the inductionhypothesis for the subspace orthogonal to vn and the set of vectors vi , i < n, andobtain the subspacesW ′
j . We define thenWi = W ′i , for i ≥ 2, and letW1 be the span
of W ′1 and vn. This proves the claim.
The claim shows that if we have m ≥ n + 2 points on the unit sphere in Rn whose
pairwise distances are at least√
2, thenm ≤ 2n, and moreover, there exist two pointsat distance precisely
√2.
• H.S.M. Coxeter: Amer. Math. Monthly 40 (1933), 192–193.• R.A. Rankin: The closest packing of spherical caps in n dimensions, Proc. Glas-
gow Math.Assoc. 2 (1955), 139–144.
Problem 3.21. A convex polygon with 2n sides has at least n diagonals not parallelto any of its sides. The equality is attained for the regular polygon.
Solution 3.21. Let l be a side of the polygon. Then we have at most n− 2 diagonalsparallel to l. In fact, let d1, d2, . . . , dp be these diagonals. Then the pair of seg-ments (d1, l) determines a quadrilateral (actually a trapezoid) having three commonsides with the polygon. Moreover, all the pairs (d2, d1), . . . , (dp−1, dp) determinequadrilaterals, each one containing two sides of the initial polygon. This implies thatp ≤ n− 2.
The total number of diagonals that are parallel to at least one side is at most2n(n− 2). But there are 2n(2n−3)
2 diagonals, and hence n(2n− 3)− 2n(n− 2) = n
diagonals with the desired property.
144 7 Geometry Solutions
Problem 3.22. Let d be the sum of the lengths of the diagonals of a convex polygonP1 · · ·Pn and p its perimeter. Prove that for n ≥ 4, we have
n− 3 < 2d
p<
[n2
] [n+ 1
2
]− 2.
Solution 3.22. Set αk = ∑ni=1 |PiPi+k|. We will prove first that
p = α1 < α2 < α3 < · · · < α[ n2 ] and α[ n2 ]+1 > · · · > αn−2 > αn−1 = p.
Let us show first that p < α2. Let AiAi+2 ∩ Ai+1Ai+3 = {Oi+1}. Observe that|Ai+1Oi+1| ≤ |Ai+1Oi+2|, since Ai+1Ai+2Oi+1 ≤ Ai+1Ai+2Oi+2. The triangleinequality in AiOiAi+1 further yields
|Ai+1Oi+1| + |Oi+1Ai+2| > |Ai+1Ai+2|.
Summing up these inequalities, and using the remark, it follows that p < α2, asclaimed.
Let now k <[n2
] − 1, and set AiAi+k ∩ Ai+1Ai+k+1 = {O}. The triangleinequality again shows that
|AiAi+k| + |Ai+1Ai+k+1| = |AiO| + |OAi+k| + |Ai+1O| + |OAi+k+1|> |Ai+1Ai+k| + |AiAi+k+1|.
By summing up over i, we obtain
2αk > αk+1 + αk−1;thus
αk+1 − αk < αk − αk−1 < · · · < α2 − α1,
and hence the claim.Moreover, if n = 2m, then
2αm > αm+1 + αm−1 = 2αm+1,
and so αm > αm−1. Thus, we have proved that αk+1 > αk for all k. By summing upall these inequalities over k, we find that 2d > (n− 3)p.
7.2 Combinatorial Geometry 145
Further, consider the triangle inequality |AiAj | < |AiAi+1| + · · · + |Aj−1Aj |and sum up over all i and j ; this yields
2d <
(([n2
] [n+ 1
2
])− 2
)p.
Problem 3.23. Find the convex polygons with the property that the function D(p),which is the sum of the distances from an interior point p to the sides of the polygon,does not depend on p.
Solution 3.23. Let K be our polygon, p an interior point, a1, . . . , an the distancesfrom p to the sides ofK , and let u1, . . . , un be unit vectors perpendicular to the sides.Choose another point, say q, in the interior of K . Denote by v the vector pq. Thedistance from q to the side with label i is therefore di = ai +〈ui, v〉, where 〈, 〉 is thescalar product. It follows that
D(q) = D(p)+⟨ n∑
i=1
ui, v
⟩.
Therefore, a necessary and sufficient condition for D(p) to be constant is that
n∑
i=1
ui = 0.
Observe that the solution can be extended to convex polyhedra in Rm.
Problem 3.24. Prove that a sphere of diameter 1 cannot be covered by n strips ofwidth li if
∑ni=1 li < 1. Prove that a circle of diameter 1 cannot be covered by n strips
of width li if∑ni=1 li < 1.
Solution 3.24. 1. The lateral area of the body obtained by cutting off the sphere withtwo planes at distance h is 2πRh. Let us assume that we have a covering by meansof the strips Bi . Then the sum of areas cut open by the strips Bi is at least the area ofthe sphere, hence π
∑ni=1 li ≥ π , contradicting our assumptions.
2. If we had such a covering, letB∗i be the spatial strip that cuts the planar stripBi ,
orthogonal to the plane of the circle. Then the union of the strips B∗i would cover the
cylinder over the disk, in particular the sphere of diameter 1. This would contradictthe first part.
Problem 3.25. Consider n points lying on the unit sphere. Prove that the sum of thesquares of the lengths of all segments determined by the n points is less than n2.
Solution 3.25. Let O be the center of the sphere and let Xi denote the given points.We denote by vj the vector
−−→OXj , and by 〈, 〉 the Euclidean inner product. We have
then|XiXj |2 = 〈vj − vi, vj − vi〉.
146 7 Geometry Solutions
Set S = ∑i,j |XiXj |2. We have then
2S =n∑
i=1
〈vi − v1, vi − v1〉 +n∑
i=1
〈vi − v2, vi − v2〉 + · · · +n∑
i=1
〈vi − vn, vi − vn〉
= 2n(|v1|2 + · · · + |vn|2)− 2∑
i,j
〈vj , vi〉
= 2nn∑
i=1
|vi |2 − 2〈v, v〉 = 2n2 − 2|v|2 ≤ 2n2,
where v = ∑ni=1 vi .
Problem 3.26. The sum of the vectors−−→OA1, . . . ,
−−→OAn is zero, and the sum of their
lengths is d . Prove that the perimeter of the polygon A1 . . . An is greater than 4d/n.
Solution 3.26. If A is an arbitrary point in the plane, then n−→OA = ∑n
i=1(−−→OAi +−−→
AiA) = ∑ni=1
−−→AiA, which implies that n|OA| ≤ ∑ |AiA|. Let us put A = Ai for
i = 1, 2, . . . , n and sum up the inequalities so obtained:
d = |OA1| + · · · + |OAn| ≤ 2
n
∑
i,j
|AiAj |.
Using the triangle inequality,
|AiAj | < |AiAi+1| + · · · + |Aj−1Aj |,we infer that
d ≤ 2
n
(p + 2p + · · · + n− 1
2p
)= n2 − 1
4np.
Therefore p ≥ 4d nn2−1
> 4dn
.
Problem 3.27. Find the largest numbers ak , for 1 ≤ k ≤ 7, with the property thatfor any point P lying in the unit cube with vertices A1 . . . A8, at least k among thedistances |PAj | to the vertices are greater or equal than ak .
Solution 3.27. At least eight distances are greater than or equal to 0, and this is sharpby taking P = Aj , and hence a8 = 0.
By taking P to be the midpoint of an edge, we get a7 ≤ 12 . Assume that a7 <
12 .
Then there exists some P for which at least two distances |PAj | are strictly smallerthan 1
2 , which is obviously false, since vertices are distance 1 apart. Thus a7 = 12 .
Consider P as the center of some face of the cube. We obtain then a6 ≤ a5 ≤√
22 .
If equality does not hold above, then there exists P for which three distances are
smaller than√
22 . However, given three vertices of the cube, there are at least two of
them at distance greater than or equal to the diagonal of a face, namely√
2. Thus our
assumption fails and so a6 = a5 =√
22 .
7.2 Combinatorial Geometry 147
Now let P be the center of the cube. We derive that a4 ≤ a3 ≤ a2 ≤ a1 ≤√
32 .
Assume that a4 <√
32 . Then there exist P and at least five distances strictly smaller
than√
32 . However, for any choice of five vertices of a cube, one finds a pair of
opposite vertices that are distance√
3 apart. This contradicts our assumptions and
thus a4 = a3 = a2 = a1 =√
32 .
Problem 3.28. The line determined by two points is said to be admissible if its slope isequal to 0, 1,−1, or ∞. What is the maximum number of admissible lines determinedby n points in the plane?
Solution 3.28. We join two points by an edge if the line determined by them is ad-missible. We obtain a graph with n vertices. By hypothesis, each vertex has degreeat most 4. Since the sum of degrees over all vertices is twice the number of edges,we find that there are at most 2n edges, thus at most 2n admissible lines. Moreover,there are only four directions in the plane, and the same computation shows that thereare at most 1
2n edges associated to every direction. This implies that for odd n, themaximum number of admissible lines is 2n− 2.
It is easy to construct examples in which we have precisely 2n admissible linesif n is even and n ≥ 12. For odd n, the associated graph is obtained by adding anisolated vertex to a maximal graph of order n− 1, and this works for any n ≥ 3.
Finally, if f (n) denotes the maximum number of admissible lines, then explicitinspection shows that f (2) = 1, f (3) = 3, and f (4) = 6, and so the upper bound2n is not attained. It can be shown that f (6) = 11 and f (10) = 19. Therefore, forall n but those from {2, 3, 4, 6, 10}, we have f (n) = 4
[n2
].
Comments 57 More generally, assume that the admissible directions are k in num-ber. The same proof shows that there therefore exist at most
[kn2
]admissible lines
determined by n points. Moreover, the minimum number of directions (not necessar-ily admissible) defined by n points is at least n− 1. See also the article:
• P.R. Scott: On sets of direction determined by n points, Amer. Math. Monthly 77(1970), 502–505.
Problem 3.29. If A = {z1, . . . , zn} ⊂ C, then there exists a subset B ⊂ A such that∣∣∣∣∑
z∈Bz
∣∣∣∣ ≥ π−1n∑
i=1
|zi |.
Solution 3.29. We reorder the numbers z1, . . . , zn,−(z1 + · · · + zn) in a sequencew1, . . . , wn+1 such that the sequence arg w1, . . . , arg wn+1 is monotonic. In thecomplex plane, the points w1, w1 + w2, . . . , w1 + · · · + wn+1, are the vertices of aconvex polygon P .
Set d(P ) for the diameter of P . Since the diameter of a polygon is the distancebetween two vertices, there exists B ⊂ A such that
d(P ) =∣∣∣∣∑
zi∈Bzi
∣∣∣∣.
148 7 Geometry Solutions
Let p(P ) be the perimeter of P . We have then
p(P ) =∑
|wi | =n∑
i
|zi | +∣∣∣∣
n∑
i
zi
∣∣∣∣.
The main isoperimetric inequality states that for any convex planar polygon Q wehave the inequality πd(Q) ≥ p(Q). Applying this to the polygon P settles the claim.
Comments 58 Using a refined isoperimetric inequality for polygons with a givennumber of vertices, one finds that there exists B ⊂ {1, . . . , n} such that
∣∣∣∣∑
i∈Bzi
∣∣∣∣ ≥ sin(kπ/(2k + 1))
(2k + 1) sin(π/2k + 1)
n∑
i=1
|zi |.
Problem 3.30. LetA1, . . . , An be the vertices of a regular n-gon inscribed in a circleof centerO. Let B be a point on the arc of circle A1An and set θ = AnOB. If we setak = |BAk|, then find the sum
n∑
k=1
(−1)kak
in terms of θ .
Solution 3.30. We have ak = 2r sin(kπn
− θ2
). Then
n∑
k=1
(−1)kak cosπ
2n=
∑(−1)kr
(sin
((2k − 1)π
2n− θ
2
)+ sin
((2k + 1)π
2n− θ
2
))
=(−1 + (−1)n+1
)r sin
(π
2n− θ
2
),
which leads us to
n∑
k=1
(−1)kak ={
2r sin(θ2 − π
2n
) (cos π
2n
)−1, for even n,
0, for odd n.
Problem 3.31. Consider n distinct complex numbers zi ∈ C such that
mini �=j |zi − zj | ≥ max
i≤n |zi |.
What is the greatest possible value of n?
Solution 3.31. Let |zn| = maxi≤n |zi |; then the zi belong to the disk of radius R =|zn| centered at the origin. Moreover, the distance between two points of complexcoordinates zi and zj is given by |zi − zj |, which by hypothesis is greater than R.In particular, we have n points whose pairwise distances are at least R within thecircle of radius R. Let O be the origin and Zi the points. We order the points Zi in
7.2 Combinatorial Geometry 149
increasing order with respect to their arguments. If the origin is among the Zi , thenwe choose it to be the first point. Further, there remain at least n − 1 points distinctfrom the origin. One knows that
n∑
i=2
ZiOZi+1 = 2π,
and thus there exists at least one angle ZiOZi+1 ≤ 2πn−1 . In particular, if n ≥ 8,
then |ZiZi+1| ≤ R
√2 − 2 cos ZiOZi+1 < R, contradicting our assumptions. Thus
n ≤ 7 and the maximal value n = 7 is reached by the configuration of six vertices ofa regular hexagon together with the origin.
Problem 3.32. The interior of a triangle can be tiled by n ≥ 9 pentagonal convexsurfaces.What is the minimal value ofn such that a triangle can be tiled byn hexagonalstrictly convex surfaces?
Solution 3.32. There is an obvious recurrence, showing how to go from k to k + 3.
We still have to find the appropriate tiling for n ∈ {9, 10, 11}. If n = 9, then wehave the following tiling of the triangle:
A slight deformation of the left corner pentagon shows that a concave quadrilateralcan also be tiled with nine pentagons. Moreover, a tiling by n pentagons of a concavequadrilateral can be adjusted as below to a tiling of a triangle with n+ 1 pentagons.This method furnishes a tiling with any number n ≥ 9 of pentagons.
150 7 Geometry Solutions
2. We will show that there are no hexagonal tilings of triangles with convex tiles.Assume the contrary. We call a tiling good if no hexagon tile has sides in commonwith (at least) two sides of the triangle.
Given an arbitrary tiling, we can construct a good tiling of a triangle as follows.First, the triangle can be assumed to be a right triangle by means of the followingtransformation. Consider the cone in R
3 over the initial tiled triangle with a conevertex far away, and cut it by a convenient plane so that one of the angles of thesection become right angled. The trace of the coned tiling in the planar section is stilla hexagonal tiling with convex polygons. Further, we double the right triangle alongone of its sides and obtain a tiling that has no hexagons with common edges with bothof the sides opposite to the doubled side. Applying again the same procedure yieldsa good tiling.
The main feature of a good tiling by convex polygons is that any hexagon havinga common side with the triangle (called a boundary hexagon) has precisely one sidein common, excepting the case in which the hexagon might have a common anglewith the triangle (called an exceptional boundary hexagon). In the last case, we havea vertex of the hexagon and its incident sides that are common with the vertex of atriangle and its incident sides. Thus there are λ ∈ {0, 1, 2, 3} exceptional boundaryhexagons. Let a∂ be the total number of boundary hexagons and ain the hexagons thatare interior, i.e., they have no sides in common with the triangle.
Construct the dual graph associated to the tiling. This is a planar graph whosevertices have degree 6. The edges going outside of the triangle will be called exterioredges of the graph, and the remaining ones interior edges. The number of vertices ofthis graph is a = ain + a∂ . Counting the sum of degrees of all vertices, we obtain6a on one side, and 2ein + eo, where ein (respectively eo) is the number of interior(respectively exterior) edges. Moreover, each exterior edge comes from a uniqueboundary hexagon, and two different edges come from different hexagons, exceptingthe exceptional hexagons, which have two outgoing edges. Thus eo = a∂ + λ. Thisimplies that ein = 3a − 1
2 (a∂ + λ).If we erase the exterior edges of our graph, then we obtain a planar polygon
partitioned into polygons (duals of the hexagons). The total number of vertices is a,the number of edges is ein and the number of faces f can be estimated from aboveas follows. We have ain vertices in the interior of a polygon with a∂ sides. Eachpolygon has at least three sides (because our initial hexagons were supposed to be
7.2 Combinatorial Geometry 151
strictly convex). Consider the sum of angles of all polygons involved. On the onehand this yields 2πain + π(aδ − 2), and on the other hand this is at least πf . Thusf ≤ 2ain + a∂ − 2. The Euler–Poincaré theorem requires that
1 = a − ein + f ≤ a −(
3a − 1
2(a∂ + λ)
)+ 2ain + a∂ − 2 = λ
2− 2 − a∂
2.
But λ ≤ 3 and a∂ ≥ 0 lead us to a contradiction.
Comments 59 The same proof shows that there does not exist a tiling of the quadri-lateral by strictly convex hexagons.
Problem 3.33. We say that a transformation of the plane is a congruence if it preservesthe length of segments. Two subsets are congruent if there exists a congruence sendingone subset onto the other. Show that the unit disk cannot be partitioned into twocongruent subsets.
Solution 3.33. Let us assume that the diskD can be partitioned asD = A∪B, whereA and B are congruent. Let p denote the center of D, and assume that p ∈ A. LetF : A → B be a congruence.
Let rs be the diameter that is perpendicular to the linepF(p). For any point x ∈ D,we have |px| ≤ 1. Thus |F(p)F (x)| ≤ 1, for any x ∈ A, and thus |F(p)y| ≤ 1,for any y ∈ B. Now |F(p)r| > 1 and |F(p)s| > 1, and thus r, s ∈ A. Fur-ther, |F(r)F (s)| = |rs| = 2 and thus F(r)F (s) is a diameter of the disk. Since|F(r)F (p)| = |F(s)F (p)| = 1, we derive that F(p) is the midpoint of the diameterF(r)F (s), and thus it coincides with the centerp. This contradiction proves the claim.
Comments 60 Actually, the same argument shows that the disk cannot be partitionedinto finitely many congruent sets. The problem is due to B.L. Van der Waerden. Moregenerally, D. Puppe showed that the same holds for any convex compact set in theplane. See also:
• H. Hadwiger and H. Debrunner: Combinatorial Geometry in the Plane, translatedby V. Klee. With a new chapter and other additional material supplied by thetranslator, New York, 1964.
Problem 3.34. Prove that the unit disk cannot be partitioned into two subsets ofdiameter strictly smaller than 1, where the diameter of a set is the supremum distancebetween two of its points.
152 7 Geometry Solutions
Solution 3.34. Observe first that the diameter of a planar set A coincides with thediameter of its closure A in R
2. Assume that the disk is partitioned into two sets Aand B of smaller diameter. The points of the boundary circle cannot belong to onlyone of the partition sets, since opposite points are at distance 1. One the other hand,the unit circle cannot be written as the union of two disjoint nontrivial closed subsets(i.e., closed intervals) because it is connected. This implies that there exists somepoint M of the unit circle that belongs to both A and B. Consider now the oppositepoint M ′ on the circle. If M ′ ∈ A, then A has diameter 1; otherwise, M ′ ∈ B and soB has diameter 1, which gives us a contradiction.
Problem 3.35. A continuous planar curve L has extremities A and B at distance|AB| = 1. Show that, for any natural number n there exists a chord determined bytwo points C,D ∈ L that is parallel to AB and whose length |CD| equals 1
n.
Solution 3.35. Assume that L does not admit chords CD parallel to AB either oflength a or of length b. We will show that L does not admit chords CD parallel toAB of length a+ b. Let Tx be the planar translation along the line AB of the numberof x units. We have, by hypothesis,
TaL ∩ L = ∅ and TbL ∩ L = ∅.This implies thatTa+bL∩TaL = ∅. On the other hand, let d(U, V ) denote the distancebetween the sets U and V in the plane, given by infP∈U,Q∈V |PQ|. Then obviouslyd(Ta+bL,L) ≤ d(TaL,L) + d(Tb+aL, TaL). This implies that d(Ta+bL,L) = 0if d(TaL,L) = d(Tb+aL, TaL) = 0. Moreover, L and Ta+b(L) are closed subsets,and thus the distance is attained, i.e., there exist P ∈ L and Q ∈ Ta+b(L) such that|PQ| = d(L, Ta+b(L)). In particular, P = Q and thus Ta+bL∩L �= ∅, proving ourclaim.
Suppose now that there is no parallel chord of length a = b = 1n
. Then we derivethat there is no such chord of length 2
n. Take a = 1
nand b = 2
n. The previous claim
shows that there is no chord of length 3n
. An easy induction shows that there is nochord of length k
nfor k ≤ n. This contradicts the fact that AB has length 1.
Comments 61 The result is due to H. Hopf, who characterized all the possible lengthsof chords arising as above.
• H. Hopf: Über die Sehnen ebener Kontinuen und die Schleifen geschlossenerWege, Comment. Math. Helv. 9 (1937), 303–319.
Problem 3.36. The diameter of a set is the supremum of the distance between twoof its points. Prove that any planar set of unit diameter can be partitioned into three
parts of diameter no more than√
32 .
Solution 3.36. The main ingredient is to show that a planar set F of unit diameter iscontained within a regular hexagon whose opposite sides are at distance 1.
Given a vector v = v1, let us consider the thinest strip Sv parallel to v thatcontains F . Then the strip Sv is bounded by two parallel lines, each line containing
7.2 Combinatorial Geometry 153
at least one point of F , by the minimality of the strip. In particular, the width of thestrip is bounded by the distance between two points of F , and so by the diameterof F .
Now let v2 be the image of v1 by a rotation of angle π3 , and let v3 be the image of therotation of angle 2π
3 . The intersection of the corresponding stripsHv = Sv1 ∩Sv2 ∩Sv3
is a semiregular (i.e., all opposite edges are parallel) hexagon having all angles 2π3 .
However, in general, this hexagon is not regular.
We choose two opposite vertices, say A and A′, of the parallelogram Sv1 ∩ Sv2 .For instance, we choose A such that the vectors spanning the parallelogram issuedfrom it are positive multiples of both v1 and v2. We denote by x the distance fromA to the strip Sv3 and by x′ the distance from A′ to the strip Sv3 . We observe thatthe hexagon Hv is regular if x = x′. Moreover, let us set f (v) = x − x′. Obviously,the function that associates to the vector v the quantity x − x′ is continuous. Let uscompute f (−v), by rotating v through π . This amounts to interchanging A with A′,and thus f (−v) = −f (v). A continuous function taking both positive and negativevalues must have a zero. Thus there exists some position of v for which f (v) = 0and so Hv is regular.
Further, the regular hexagon of width 1 can be partitioned into three pieces having
diameter√
32 , as in the figure below:
This proves the claim.
Comments 62 K. Borsuk proved in 1933 that a planar set of diameter 1 can bepartitioned into subsets of diameter d < 1. The main ingredient above is a lemmadue to J. Pál, from 1920. B. Grünbaum proved (improving on ideas of D. Gale) that aset of unit diameter in R
3 is contained in a regular octahedron of width 1, with three ofits vertices (one vertex on each diagonal) cut off by planes orthogonal to the diagonals,
154 7 Geometry Solutions
at distance 12 from the center. Using a suitable dissection of this polyhedron, one finds
that the set F can be partitioned into four pieces having diameter not bigger than√
6129030 − 937419√
3
1518√
2≈ 0.9887.
Gale’s conjecture from 1953, that one can always get a partition into four pieces of
diameter no bigger than√
3+√3
6 , is still open.
• D. Gale: On inscribing n-dimensional sets in a regular n-simplex, Proc. Amer.Math. Soc. 4 (1953), 222–225.
• B. Grünbaum: A simple proof of Borsuk’s conjecture in three dimensions, Math.Proc. Cambridge Philos. Soc. 53 (1957), 776–778.
Problem 3.37. 1. Prove that a finite set of n points in R3 of unit diameter, can be
covered by a cube of side length 1 − 23n(n−1) .
2. Prove that any planar set of n points having unit diameter can be partitioned into
three parts of diameter less than√
32 cos 2π
3n(n−1) .
Solution 3.37. 1.A direction will mean below a line without specified orientation. Theangle between two directions is the smallest among the four angles they define, andthus it is always within the range
[0, π2
]. Consider the set of n points {x1, x2, . . . , xn},
which determine at most N = n(n−1)2 distinct directions xixj , which we denote
by l1, l2, . . . , lN . We want to prove first that there exist three mutually orthogonaldirections y1, y2, y3, so that
min1≤i≤3,1≤s≤N ∠(yi, ls) ≥ arccos
(1 − 1
3N
).
Consider an orthogonal frame in R3 given by three mutually orthogonal vectors
e1, e2, e3, and choose the rotation R1 and R2 of the space with the property thatR1e1 = e2, R1e3 = e3, R2e1 = e3, R2e2 = e2. Using these rotations, we have
min1≤i≤3,1≤s≤N ∠(yi, ls) = min
1≤s≤N min(∠(y1, ls),∠(y1, R1ls),∠(y1, R2ls)).
Consider the set of 3N directions D = {lj , R1lj , R2lj , 1 ≤ j ≤ N}. We will provethe following estimate from below:
maxy
minl∈D ∠(y, l) ≥ arccos
(1 − 1
3N
)= θ,
for an arbitrary set D of 3N directions. Assume the contrary. The set of directionsy making an angle smaller than θ with a given direction l form a symmetric coneCl,θ of angle 2θ with axis l. If there is no direction y satisfying the inequality above,then the union of all cones Cl,θ , over l ∈ D will cover the set of all directions inspace. Each symmetric cone Cl,θ intersects the unit sphere along a symmetric pair of
7.2 Combinatorial Geometry 155
spherical caps of radius θ . In particular, these spherical caps cover the surface of thewhole sphere, since a point not covered yields a direction not belonging to the union.Furthermore, this implies that the total area of the spherical caps is (strictly) greaterthan the area of the sphere. Now, the area of each spherical cap is 2π(1 − cos θ), andthus the total area covered by the cones is 12πN(1− cos θ) = 4π . But 4π is the totalarea of the sphere, which contradicts the claim. Observe that the inequality must bestrict, since N ≥ 1 and any covering of the sphere by nontrivial spherical caps willhave some overlaps.
Consider now three mutually orthogonal directions yj satisfying the claim above.Let further ξj ⊂ yj be the image of the given set of points under orthogonal projectionontoyj . The distance between the images ofxi andxk onyj is |xixk| cos ∠(yj , xixk) ≤sin θ . Thus each subset ξj is contained in some interval of length less than cos θ , andthus the set of points is contained in the strip of width sin θ , orthogonal to yj . Theintersection of the three orthogonal strips associated to the direction yj is a cube ofside length cos θ , and the claim follows.
2. The second part follows using the same idea. This time, we consider threedirections y1, y2, y3 in the plane such that y2 is obtained by a counterclockwiserotation R1 of angle π
3 from y1, and y3 by a counterclockwise rotation R2 of angle π3
from y2. We claim that for any N directions in the plane, we have
min1≤i≤3,1≤s≤N ∠(yi, ls) ≥ π
6N= α.
The same trick as above shows that the 3N symmetric angle sectors of angle 2α coverthe set of all directions and thus the unit circle. Now, the total length of arcs coveredby these symmetric cones is 12Nα = 2π , and the claim follows.
Therefore, the set of points is contained within the intersection of three strips of
width w = cos(
2π3n(n−1)
), which are orthogonal, respectively, to the directions yj .
The intersection is a semiregular hexagon H . Since all three strips have the samewidth, we find that there are only two different side lengths, as in the figure below:
This means that H is the intersection of two equilateral triangles of the sameincenter O. Let A1, . . . , A6 be the vertices of H . We have then |A1A2| = |A3A4| =|A5A6| = a and |A2A3| = |A4A5| = |A6A1| = b. Moreover, by computing the
156 7 Geometry Solutions
width of H , we find that the lengths a, b are constrained to satisfy a + b = 2√3w.
Assume from now on that a ≤ b.We divide the hexagonH by means of three segmentsOP,OQ,OR orthogonal to the three longest sides of the hexagon. The diameter ofthe pentagonOPA1A2Q is one of its diagonals, and one computes easily |PQ| = a+b2 , |OA1| =
√3b2+(2a+b)2
2√
3, and |PA2| ≤ |PQ| since ∠(PA2Q) ≥ ∠(PQA2) = π
3 .
Using the relation between a and b, we obtain that |PQ| ≤ 34 (a + b) ≤
√3
2 w, withequality only when a = b and the hexagon is regular. On the other hand, |OA1| ≤ 2w
3 ,with equality when a = 0. This implies that the diameter of each piece is at most√
32 w, where w has the value found above.
Comments 63 This version of Borsuk’s problem for finite sets and its higher-dimensional generalization for covering by cubes was given in:
• L. Funar: A minimax problem in geometry (in Romanian), Proc. 1986 Nat. Conf.Geometry and Topology - Tîrgoviste, Univ. Bucharest, 91–96, 1988.
Problem 3.38. Prove that any convex body in Rn of unit diameter having a smooth
boundary can be partitioned into n+ 1 parts of diameter d < 1.
Solution 3.38. The first step is to observe that the claim holds for the unit ball Bn.We consider the regular spherical simplex obtained by projecting the regular simplexonto the sphere from its incenter. The n + 1 faces of the spherical simplex form adissection into pieces of diameter smaller than 1, since there are no opposite pointslying in the same face. Otherwise, one can compute explicitly the diameter of eachpiece.
Consider the convex body F with smooth boundary W ⊂ Rn. The only point
where the smoothness is used is when we assume that there exists a unique tangenthyperplane TpW at each point p of W . Moreover, there exists a unique unit vectorvp in R
n, based at p, that is orthogonal to the tangent hyperplane TpW and pointstoward the half-space that does not contain F . This permits us to define a smoothmap G : W → Sn−1 to the unit sphere centered at the origin O, usually called theGauss map in differential geometry. We associate to the point p ∈ W the unique pointG(p) ∈ Sn−1 for which the vectors OG(p) and vp are parallel, and so the tangentspace TpW is parallel to the tangent hyperplane TG(p)Sn−1 at the point G(p) on thesphere. Since F is convex andW is smooth, the mapG is a smooth diffeomorphism.
Consider now the preimage by G of a partition of the unit sphere into pieces ofdiameter smaller than 1. This yields a partition of W into n + 1 pieces, which wedenote by A1, A2, . . . , An+1. We claim that each piece Aj has diameter smaller than1. Assume the contrary. We can replace the Aj by their closures, without modifyingtheir diameters. Now there should exist two points x, y ∈ Aj at distance 1, which isthe diameter of F . Let TxW and TyW be the tangent planes atW at x, y respectively.ThenTxW andTyW are orthogonal to the segment xy. Otherwise, a small perturbationx′ ∈ W of x in some tangent direction from TxW , making an obtuse angle with xy,will increase the length of |x′y|, but the diameter of W is no larger than 1.
7.2 Combinatorial Geometry 157
Therefore G(x) and G(y) are opposite points on the unit sphere, and moreover,they belong to the same set of the partition, but this is impossible, since the pieces ofthe partition are of diameter strictly smaller than 1.
Comments 64 The question whether any set F in Rn can be partitioned into n+ 1
pieces of strictly smaller diameter is known as Borsuk’s problem. Since the diameterof a set equals the diameter of its convex hull, it suffices to consider the problem theconvex sets. The result in the problem above was obtained by H. Hadwiger in 1945.The case left open is that in which the convex body has corners.
There was no significant progress in the subject until 1993, when J. Kahn and G.Kallai gave an unexpected counterexample by showing that the minimal numberf (n) of pieces needed for a dissection into parts of strictly smaller diameter satisfiesf (n) ≥ (1.1)
√n for large n and that f (n) > n + 1 for n ≥ 1825. The subsets that
they considered are finite sets of points. This bound was subsequently lowered by N.Alon and then, by A. Hinrichs, and C. Richter to n = 298. A thorough treatmentof the subject is in the recent book of Boltyanski, Martini, and Soltan. However, the4-dimensional case is still open.
• V. Boltyanski, H. Martini, and P.S. Soltan: Excursions into Combinatorial Geom-etry, Springer, Universitext, 1997.
• H. Hadwiger: Überdeckung einer Menge durch Mengen kleineren Durchmesser,Commentarii Math. Helvet. 18 (1945/1946), 73–75, 19 (1946/1947), 71–73.
• A. Nilli: On Borsuk’s problem, Jerusalem combinatorics ’93, 209–210, Contemp.Math., 178, Amer. Math. Soc., Providence, RI, 1994.
• J. Kahn and G. Kalai: A counterexample to Borsuk’s conjecture, Bull.Amer. Math.Soc. 29 (1993), 1, 60–62.
Problem 3.39. Let D be a convex body in R3 and let σ(D) = supπ area(π ∩ D),
where the supremum is taken over all positions of the variable plane π . Prove thatDcan be divided into two parts D1 and D2 such that σ(Di) < σ(D).
Solution 3.39. Choose a point in the interior ofD and a sequence of chords l1, l2, . . . ,ln, . . . of D passing through this point, the chords being dense in D. Choose also asequence ε of positive numbers decreasing rapidly to zero. Let Cn be the set ofpoints at distance less than εn from ln. By choosing ε small enough, we can ensurethat σ(Cn) < 2−nσ (D). Set D1 = ∪∞
n=1Cn and D2 = D − D1. Then σ(D1) <∑∞n=1 σ(Cn) < σ(D). It is also clear that for any plane π , area(π ∩ D2) < σ(D).
But D2 is a compact and area(π ∩D2) is a continuous function on the plane π . Theset of planes that intersect D can be parameterized by a point in D (which will beconsidered the origin) times the projective space RP 2, and thus by a compact set.Thus area(π ∩ D2) achieves its minimum, which will therefore be strictly smallerthan σ(D).
Comments 65 The problem was proposed by L. Funar and the solution above wasgiven by C.A. Rogers. A more interesting question is to consider dissections into convexsubsets. The additional convexity assumption implies that the boundary structure of
158 7 Geometry Solutions
each piece (at the interior of the body) should be polyhedral. In particular, the minimalnumber of convex pieces with smaller σ , which the unit ball requires, is 3.
If the bodyD is strictly convex (meaning that the open segment determined by two ofits boundary points is contained in the interior of D) or D is a polyhedron withoutparallel edges, then we can show that four pieces suffice. By a recent result of M.Meyer, two plane sections of maximal area cannot be disjoint. Fix a maximal-areacross-section F and let F(ε) be the set of points of D, at distance less than ε fromF . The complement of F(ε) has two components, A+ and A−, which are of strictlysmaller σ , because there is no maximal-area plane section contained completely inD − F(ε). Further, F(ε) can be subdivided, by a hyperplane orthogonal to F , intotwo pieces of smaller σ if ε is small enough. However, we don’t know whether threepieces suffice for an arbitrary convex body.
A related problem is whether we have σ(D) ≤ 14 area(∂D), where ∂D denotes the
boundary of D. This would imply that the number of maximal area planar sectionswith disjoint interiors is at most 4, with equality for the regular simplex.
• L. Funar: Problem E 3094, Amer. Math. Monthly 92 (1985), 427.• M. Meyer: Two maximal volume hyperplane sections of a convex body generally
intersect, Period. Math. Hungar. 36 (1998), 191–197.
Problem 3.40. If we have k vectors v1, v2, . . . , vk in Rn and k ≤ n + 1, then there
exist two vectors making an angle θ with cos θ ≥ − 1k−1 . Equality holds only when
the endpoints of the vectors form a regular (k − 1)-simplex.
Solution 3.40. Assume that the vectors vj are unit vectors. Then
∑
1≤i<j≤k|vi − vj |2 ≤
⎛
⎝∑
1≤i<j≤k|vi − vj |2
⎞
⎠ +∣∣∣∣∣
k∑
i=1
vi
∣∣∣∣∣
2
= k
k∑
i=1
|vi |2 = k2.
Thus there exists at least one pair i, j for which |vi − vj |≤ 2kk−1 . This is equivalent to
the fact that the angle θ between vi and vj satisfies the claimed inequality.
Problem 3.41. 1. Consider a finite family of bounded closed convex sets in theplane such that any three members of the family have nonempty intersection.Prove that the intersection of all members of the family is nonempty.
2. A set of unit diameter in R2 can be covered by a ball of radius
√13 .
Solution 3.41. 1. Assume that the convex sets are A1, A2, . . . , Ak and k ≥ 4. Weclaim that if each k − 1 sets among them have a common point, then all of themshould have a common point. The result of the problem follows by using recurrentlythe previous claim.
Before proceeding, we observe that given a set of k ≥ 4 points in the plane, wecan find two subsets whose convex hulls intersect nontrivially. Here the convex hullof a set of points is the smallest convex polygon containing these points (and thus all
7.2 Combinatorial Geometry 159
segments pairwise connecting them). In fact, if we choose four points among them,then their convex hull is either a triangle containing the fourth point in its interior, orelse a quadrilateral having two diagonals crossing each other.
Let pi be a point belonging to ∩j �=i,j≤kAi and possibly not to Ai . Thereexists therefore a partition of the set {p1, p2, . . . , pk} = P ∪ Q, where P ={pi1 , pi2 , . . . , pim} and Q = {pj1 , pj2 , . . . , pjk−m}, such that the convex hull of thesubset P intersects in at least one point the convex hull of the subset Q. Recall thateach point ofP belongs to ∩i �∈{i1,i2,...,im}Ai , which is convex as the intersection of con-vex sets. Thus the convex hull of P is equally contained within ∩i �∈{i1,i2,...,im}Ai . Thesame argument shows that the convex hull of Q is contained in ∩j �∈{j1,j2,...,jk−m}Aj .Since the two convex hulls have at least one common point, this therefore belongs to∩ki=1Ai , as claimed.
2. It suffices to see that any finite subset is contained in a disk of radius√
13 , which
means that the intersection of balls centered at that points of radius√
13 is nonempty.
By the previous claim, it suffices to consider the case of three points. If the trianglemade by these points is obtuse, then we can take the disk whose diameter is the largestside of the triangle. If not, then all angles of the triangle are acute and there exists at
least one angle, say A ≥ π3 . The circumradius of the triangle is R = a
2 sinA ≤√
13 .
Comments 66 The first part of the problem is the particular case of the celebratedHelly’s theorem, proved in 1923: if a family of closed convex sets in R
n has theproperty that any n + 1 among them have a common point, then the intersection ofall sets is nonempty.
The second claim was generalized by Jung in 1901, who proved that any set of diameter1 in R
n admits an unique n-ball of minimal radius covering it, and the minimal radius
is bounded by√
n2(n+1) . The original proof of Jung’s theorem was quite complicated,
but there exists an elementary proof given later by Blumenthal and Wahlin.
It is worthy of notice that any subset of Rn of diameter 1 can be covered also by a
simplex of diameter√n(n+1)
2 , according to Gale.
• L.M. Blumenthal, G.E. Wahlin: On the spherical surface of smallest radius en-closing a bounded subset of n-dimensional Euclidean space, Bull. Amer. Math.Soc. 47 (1941), 771–777.
• L. Danzer, B. Grünbaum, andV. Klee: Helly’s theorem and its relatives, Convexity,Proc. Symposia Pure Math. vol. 7, AMS, 1963, 101–180.
• D. Gale: On inscribing n-dimensional sets in a regular n-simplex, Proc. Amer.Math. Soc. 4 (1953), 222–225.
• H.W.E. Jung: Über die kleinste Kugel, die eine räumliche Figur einschliesst, J.Reine Angew. Math. 123 (1901), 241–257.
Problem 3.42. Let T be a right isosceles triangle. Find the disk D such that thedifference between the areas of T ∪D and T ∩D is minimal.
160 7 Geometry Solutions
Solution 3.42. Let us show that the circle C boundingD has the following two prop-erties, valid for an arbitrary triangle T :
1. The circle C is split into three arcs outside T and three arcs inside T . Then thetotal length of the outside arcs equals the total length of the inside arcs.
2. The three chords determined by the edges of T have their lengths proportional tothe respective side lengths.
1.Assume thatC has fixed center.When the radius ofC increases by ε, area(T−D)decreases as linε
2, where lin is the length of the inside arcs. On the other hand,area(D − T ) decreases as louε
2, where lou is the length of the outside arcs. Thus wehave a local minimum only if (lin − lou)ε
2 does not increase and thus lin = lou.2. Assume now that C has fixed radius. Since area(T − D) − area(D − T ) =
area(T )− area(D), it is sufficient to minimize area(D−T ). Consider that the centerof D is slightly translated by εu, where u is a unit vector.
Then we can compute the change of area(S − T ) by summing the contributionof each chord sector. The contribution of A1A2 is εu × (
A1A2)
and thus the totalvariation is
εu× (A1A2 + B1B2 + C1C2
).
The variation must be zero in all directions u and so we obtain the condition
A1A2 + B1B2 + C1C2 = 0.
This implies that the chords are proportional to the side lengths.
If T is right isosceles triangle of unit side length, then the centerO of the circleCbelongs to the altitude, by symmetry considerations. Denote by x the distance fromthe centerO to the edges CB and CA, by y the distance fromO to the edgeAB. Theangles determined by the outside chords are 2α, 2β, 2β respectively. SinceO belongsto the altitude, we have y + √
2x = 12 . Moreover, the first claim above implies that
α + 2β = π2 and thus cosα = sin 2β. This translates into y
r= 2x
r2
√r2 − x2 and
thus y2r2 = 4x2r2 − 4x4. The second statement above implies that√r2 − y2/
√2 =√
r2 − x2, and thus introducing above, we obtain 2r2 = 4x − 1. In particular, xsatisfies the equation
16x4 − 16x3 − 12x2 + 8x − 1 = (4x2 + 2x − 1
)(4x2 − 6x + 1
) = 0.
7.2 Combinatorial Geometry 161
The only convenient root is x =√
5−14 , since among the remaining ones, one is
negative, another one is larger than 1, and the last one would lead to a negative valuefor r . Thus
r =√
1
2
(√5 − 2
), y = 1 + √
2 − √5
2√
2,
and the circle is determined by these parameters.
Problem 3.43. Let r be the radius of the incircle of an arbitrary triangle lying in the
closed unit square. Prove that r ≤√
5−14 .
Solution 3.43. It is clear that the inradius of a triangle contained in a larger triangleis not bigger than the inradius of the bigger triangle. Thus it suffices to consider onlytriangles whose vertices lie on the boundary of the square. Further, at least one vertexof the triangle can be assumed to be in a corner of the triangle. If two vertices ofthe triangle belong to the same edge, then we can assume that these vertices are theendpoints of that edge, as above. Thus it suffices to consider the triangleOAB whosevertices are O = (0, 0), A = (a, 1), and B = (1, b) with 0 ≤ a, b ≤ 1. The inradiusof a triangle of area S and semiperimeter p is given by r = S
p. Therefore,
r = 1 − ab√1 + a2 + √
1 + b2 + √(1 − a)2 + (1 − b)2
.
The inequality to be proved is then equivalent to
F(a, b) =√
1 + a2 +√
1 + b2 +√(1 − a)2 + (1 − b)2 − (√
5 + 1)(1 − ab) ≥ 0
for 0 ≤ a, b ≤ 1. We have F(a, 0) ≥ 0. Further, ∂2F∂2a
> 0 by a simple computation.
Let ε2 = 1 − 6√
3(1+√
5)2.
1. If 1 ≤ b ≤ 14 , set f (x) = F(x, b). Then we observed above that f ′(x) is
increasing, and since f ′(0) = (1 + √
5)y − 1
y2−2y+2> 0, then f ′(x) > 0 for any
x, and thus f (x) is increasing. In particular, F(a, b) ≥ F(0, b) ≥ 0, provided that14 ≤ b ≤ 1.
2. If 0 ≤ a ≤ 14 and 0 ≤ y ≤ ε, we have
f ′(x) = (1 + √
5)b + x√
1 + x2− 1 − x
√(1 − x)2 + (1 − b)2
≤ (1 + √
5)ε + x√
1 + x2− 1 − x√
x2 − 2x + 2.
Thus f ′( 14 ) < 0. As f ′ is increasing, we have f ′(x) < 0, for x ∈ [
0, 14
], and so f
is decreasing on[0, 1
4
]. But f
( 14
) = F( 1
4 , b) = F
(b, 1
4
) ≥ 0, by the previous case.Thus F(a, b) ≥ 0, for 0 ≤ a ≤ 1
4 and 0 ≤ y ≤ ε.3. Since F is symmetric, it suffices to check now the case ε ≤ a ≤ 1
4 andε ≤ b ≤ 1
4 . Actually, we will use only the fact that ab ≥ ε2. This amounts to saying
162 7 Geometry Solutions
that the area S of the triangle OAB is S = 1−ab2 ≤ 1−ε2
2 . Any triangle satisfies the
inequality S ≤ p2
3√
3, obtained from the Heron formula for the area by the means
inequality. In our case, we have
r = S
p≤ S
√3√
3S≤ √
5 + 1.
This settles our claim.
Comments 67 The solution presented here is due to Abi-Khuzam and Barbara:
• L. Funar: Problem 6477, Amer. Math.Monthly 81 (1984), 588.• F.Abi-Khuzam, R. Barbara: A sharp inequality and the inradius conjecture, Math.
Inequalities and Appl. 4 (2001), 323–326.
Problem 3.44. Let P be a point in the interior of the tetrahedron ABCD, with theproperty that |PA| + |PB| + |PC| + |PD| is minimal. Prove that APB = CPD
and that these angles have a common bisector.
Solution 3.44. Let fA(x) = |AX|. Then fA : E3 → E3 is a smooth function onE3 − {A}. Moreover, the gradient function (∇fA)(X) is the unit vector of directionAX, namely AX
|AX| . Consider now the function g = fA + fB + fC + fD for P ∈int(ABCD). Then g having a minimum at P implies that
(∇g)(P ) =∑
(∇fA)(P )+∑
(∇fB)(P )+∑
(∇fC)(P )+∑
(∇fD)(P ) = 0.
If a, b, c, d are the unit vectors of directions AP,BP,CP,DP , then the previousconditions reads a+ b+ c+ d = 0, or alternatively, −(a+ b) = c+ d. Observe nowthat (a+b) is the direction of the bisector of the angle APB, while c+d is the directionof the bisector of CPD. Thus these two angles have a common bisector. On the otherhand, we have 〈a + b, a + b〉 = 〈c + d, c + d〉, and because |a| = |b| = |c| = |d|,by simplifying the terms, we obtain that 〈a, b〉 = 〈c, d〉. This is equivalent to sayingthat APB = CPD.
Problem 3.45. Let OA1, . . . , OAn be n linearly independent vectors of lengthsa1, . . . , an. We construct the parallelepiped H having these vectors as sides. Thenconsider the n altitudes in H as a new set of vectors and further, construct the paral-lelepiped E associated with the altitudes. If h is the volume of H and e the volumeof E, then prove that
he = (a1 . . . an)2.
Solution 3.45. Let vi = OAi and wi be the altitude vectors. We have the relations
〈wj , vi〉 = δij‖vi‖2,
where δij is Kronecker’s symbol and 〈, 〉 is the scalar product. The linear functionalsfi(vj ) = δij‖vi‖2 are linearly independent, and thus the dual vectors wj are linearlyindependent. Thus E is well defined.
7.2 Combinatorial Geometry 163
Moreover, let A be the matrix consisting of the components of the vi’s and let Bbe the matrix consisting of the components of the wj ’s. It follows that
AB� = diag(‖v1‖2, . . . , ‖vn‖2),
where diag(c1, . . . , cn) denotes the diagonal matrix with given entries. Therefore
vol(E)vol(H) = | det A| · | det B�| = (a1 . . . an)2.
Problem 3.46. Let F be a symmetric convex body in R3 and let AF,λ denote the
family of all sets homothetic to F in the ratio λ that have only boundary points incommon with F . Set hF (λ) for the greatest integer k such that AF,λ contains k setswith pairwise disjoint interiors. Prove that
hF (λ) ≤ (1 + 2λ)3 − 1
λ3 .
Solution 3.46. Set Bλ = ∪H∈AF,λH . We prove first that
Bλ ⊂ (1 + 2λ)F,
where αF denotes the homothety of F in ratio α. If v is a point of the boundary ∂F ,let v′ be given by |ov′| = (1 + λ)|ov|, where o is the symmetry center of F . SetFv for the set in AF,λ having center at v′. Let x be a point on the boundary of Fv ,ox ∩ ∂F = {a}, ov ∩ ∂Fv = {q}, and let vx′′ be parallel to qx, with x′′ ∈ ∂F . Then
qv′x = v′xo+ v′ox ≥ v′ox,
so thatvox ≤ vox′′ = qv′x.
Since F is convex, the intersection of the two segments |oa| ∩ |vx′′| is nonempty andcontains at least one point, say b. Then |vx′′| ⊂ F , b ∈ F , b ∈ |ox|. Since
‖oa‖‖ox‖ ≥ ‖ob‖
‖ox‖ = ‖ov‖‖oq‖ = 1
(1 + 2λ),
the point x belongs to (1 + 2λ)F .Now assume that we have k subsetsFi ∈ AF,λ that have pairwise disjoint interiors.
Then their union is contained in Bλ, and all of them are disjoint from F . This meansthat their union is contained in (1 + 2λ)F − F . Their total volume cannot thereforeexceed vol((1 + 2λ)F − vol(F ), and thus their number is bounded by
1
vol(F )
(vol(1 + 2λ)F − vol(F )
) = (1 + 2λ)3 − 1
λ3 .
Comments 68 Hadwiger considered the problem of finding estimates for hF (1)(which is called the Hadwiger number of F ) and proved in 1957 that
164 7 Geometry Solutions
n2 + n ≤ hF (1) ≤ 3n − 1
for a convex body F in Rn. Grünbaum conjectured in 1961 (and proved for n = 2)
that for any r satisfying these inequalities there exists a convex body F such thathF (1) = r .
More generally, the proof used above shows that
hF (λ) ≤ (1 + 2λ)n − 1
λn.
We have equality above only if 1λ
∈ Z+ and F is a parallelohedral body. The resultis due to V. Boju and L. Funar.
There are no good estimates from below except in dimension 2. In fact, if n = 2,the function λhF (λ) approaches an interesting invariant of the oval F , called theintrinsic perimeter. Define the norm ‖ ∗ ‖F by means of
‖xy‖F = ‖xy‖‖oz‖ ,
where z is a point of the boundary ∂F such that xy and oz are parallel. This iscalled the Minkowski norm (or metric) defined by the oval F . The intrinsic perimeterp(F) of ∂F is the limit of the perimeters of polygons inscribed in F that approach∂F . For instance, the intrinsic perimeter of a circle is 2π and that of a square is 8.Golab in 1932 and Reshetnyak in 1953 showed that the intrinisc perimeter satisfiesthe inequalities
6 ≤ p(F) ≤ 8
with equality in the left side only when F is a hexagon, and on the right side onlywhen F is a rectangle.
Then it was proved by V. Boju and L. Funar that the intrinisc perimeter is relatedto Hadwiger numbers by means of the formula
p(F) = 2 limλ→0
λhF (λ),
and moreover, we have
3 + 3
λ≤ hF (λ) ≤ 4 + 4
λ,
with equality on the left side only if 1λ
∈ Z+ and F is an affine regular hexagon.More about packing and covering invariants of convex bodies can be found in the
recent book of Böröckzy, and a complete survey on the Minkowski geometry in thebook of Thompson.
• V. Boju and L. Funar: Generalized Hadwiger numbers for symmetric ovals, Proc.Amer. Math. Soc. 119 (1993), 931–934.
• K. Böröczky Jr.: Finite packing and covering, Cambridge Tracts in Mathematics,154, Cambridge University Press, Cambridge, 2004.
• H. Hadwiger: Über Treffanzahlen be. translationsgleichen Eikörpern, ArchivMath. 8 (1957), 212–213.
7.2 Combinatorial Geometry 165
• A.C. Thompson: Minkowski geometry, Encyclopedia of Mathematics and Its Ap-plications, 63. Cambridge University Press, Cambridge, 1996.
Problem 3.47. Let � denote the square of equations |xi | ≤ 1, i = 1, 2, in the plane,and let A = (a1|a2) be an arbitrary nonsingular 2 × 2 matrix partitioned into twocolumns. We identify each column with a vector in R
2. Prove that the followinginequality holds:
minA
maxx∈�
∣∣∣∣〈a1, x〉〈a2, x〉
det A
∣∣∣∣ = 1
2,
where 〈x, y〉 = x1x2 + y1y2 is the usual scalar product.
Solution 3.47. Let f (A, x) be the function under minimax. Taking A0 = ( 1 1−1 1
),
we obtain
maxx∈� |f (A0, x)| = 1
2.
Thus it suffices to show that for general A, we have
maxx∈� |f (A, x)| ≥ 1
2.
This is equivalent to showing that
| det A| ≤ 1, provided that maxx∈� |〈a1, x〉〈a2, x〉| = 1
2.
Assume that the contrary holds, and so | detA| > 1. The set of points in the plane withcoordinates (〈a1, x〉, 〈a2, x〉) ∈ R
2, for x ranging in the square�, is a parallelogramπ centered at the origin O and having area 4| detA| > 4.
According to our claim, the parallelogram π is contained in the planar regionZ = {
(u, v) ∈ R2; |uv| ≤ 1
2
}. Let us consider the point P = (0, s), where s > 0, is a
boundary point of π , and letL ⊂ R2 be the lattice generated by the vectors (1/s, s/2)
and (−1/s, s/2). The lattice has area 12 . Then, according to Minkowski’s theorem,
the interior of π contains at least one point of the lattice L, different from the origin.Let this point be given by
m(1/s, s/2)+ n(−1/s, s/2) = ((m− n)/s, (m+ n)s/2),where m, n ∈ Z.
We can also suppose gcd(n,m) = 1. Since the interior of π is contained in Z, itfollows that |m2 − n2| < 1 and thus m2 = n2. The only possibilities are (0,±s) and(±2/s, 0). The points (0,±s) are on the boundary of π and they are not convenient.If Q = (2/s, 0) ∈ int(π), then the midpoint S of the segment |PQ| should alsobelong to the interior of π . However, S = (1/s, s/2) belongs to the boundary curveof equation |uv| = 1/2 of the region Z containing π , so it cannot lie within theinterior of π . This contradiction proves our claim.
Problem 3.48. We denote by δ(r) the minimal distance between a lattice point andthe circle C(O, r) of radius r centered at the originO of the coordinate system in theplane. Prove that
limr→∞ δ(r) = 0.
166 7 Geometry Solutions
Solution 3.48. LetG be a lattice point on the axisOx such that |OG| = [r]. Draw aperpendicular line atG toOx that intersects the circleC(O, r) at the pointC situatedbetween the lattice points A and B, in the upper half-plane. We choose B to be thepoint that lies in the interior of the disk D(O, r) of radius r . Notice that |AB| = 1,because these are neighboring lattice points on the same axis line.
Let us now draw BD⊥AB, where D is again a lattice point, |BD| = 1, and D issituated outside the disk D(O, r).
Since the problem asks for the value at the limit of δ(r) when r → ∞, we canassume that r > 4.
Let now consider the line CF that is tangent to the circle C(O, r) and such thatF ∈ BD. We show first that |BF | ≤ |BD|.
Assume the contrary, namely that |BF | > |BD|. Then CFB < CDB ≤ ADB =π4 . Furthermore,
sin CFB = sin OCG = sin
(arcsin
[r]
r
)>r − 1
r>
√2
2= sin
π
4, if r > 4.
Thus CFB > π4 , which is false.
Consequently, |BF | ≤ |BD| = 1. The triangles CBF andOCG are similar, andtherefore
|BC||OG| = |CG|
|OG| =√r2 − [r]2
2<
√r2 − (r − 1)2
r<
√2r
r=
√2√r.
Now it is clear that
δ(r) ≤ |BC| ≤ |BC||BF | =
√2√r, for r > 4.
Thus limr→∞ δ(r) = 0.
Problem 3.49. Consider a curve C of length l that divides the surface of the unitsphere into two parts of equal areas. Show that l ≥ 2π .
Solution 3.49. Let C′ be the symmetric (opposite, or antipodal) of C with respect tothe centerO of the sphere. If C∩C′ = ∅, then int C∩ int C′ = ∅ and thus the area ofthe sphere is strictly greater than the sum of areas of int C and int C′. By hypothesis,the latter is just the area of the sphere, which is a contradiction.
Therefore, C ∩C′ contains at least one point, say P . The point P ′, the symmetricofP with respect to the symmetry center ofC∩C′, should also belong toC∩C′, sincethe later is symmetric. Finally, note that the shortest curve on the sphere that joinstwo opposite points is an arc of a great circle, and since C is made of two differentarcs joining P and P ′, it follows that l ≥ 2π .
Comments 69 More generally, let K be a body with a symmetry center O and setr = minP∈∂K |OP |, where ∂K denotes the boundary surface of K . If C ⊂ ∂K isa curve that separates two regions on ∂K of the same area, then the length l of Csatisfies l ≥ 2πr .
7.2 Combinatorial Geometry 167
Problem 3.50. Let K be a planar closed curve of length 2π . Prove that K can beinscribed in a rectangle of area 4.
Solution 3.50. Let bK(θ) be the width of the figure K in the direction θ , identifiedwith a vector θ ∈ S1. If R(K, θ) denotes the area of the rectangle circumscribing Kand having one side parallel to θ , then
R(K, θ) = bK(θ)bK
(θ + π
2
).
Thus
minθR(K, θ)
12 ≤ 1
2π
∫ 2π
0bK(θ)
12 bK
(θ + π
2
)1/2dθ,
with equality holding iff bK(θ)bK(θ + π
2
)is constant. From the Cauchy–Schwartz
inequality, one derives
minθR(K, θ) ≤ 1
4π2
(∫ 2π
0bK(θ) dθ
)2
,
the equality holding iff K is a curve of constant width. We are able now to apply thefollowing theorem of Cauchy, which expresses the length L(K) of the curve K interms of the integral of its width with respect to all directions:
L(K) = 1
2
∫ 2π
0bK(θ) dθ.
This implies that
minθR(K, θ) ≤ 1
π2L(K)2 ≤ 4,
where the second inequality uses the hypothesis L(K) ≤ 2π .
Comments 70 We can consider the parallelograms with a given acute angleφ insteadof rectangles. The same argument shows that
minθRφ(K, θ) ≤ cosφ
π2 L(K)2.
See also:
• E. Lutwak: On isoperimetric inequalities related to a problem of Moser, Amer.Math. Monthly 86 (1979), 476–477.
Problem 3.51. 1. Consider a family of plane convex sets with area a, perimeter p,and diameter d. If the family covers the areaA, then there exists a subfamily withpairwise disjoint interiors that covers at least area λA, where λ = a
a+pd+πd2 .2. Assume that any two members of the family have nonempty intersection. Prove
that there exists then a subfamily with pairwise disjoint interiors that covers areaat least μA, where μ = a
πd2 .
168 7 Geometry Solutions
Solution 3.51. 1. If K is a member of the family and K(d) the set of points at dis-tance less than d from K , then it is known that area(K(d)) = a + pd + πd2. Let{K1,K2, . . . , Kn} be a maximal subfamily with disjoint interiors. Then every mem-ber of the family intersects someKi , and thus it is contained in someKi(d). Thus thefamily is contained within
⋃ni=1Ki(d). Thus the area covered by all members of the
family is at most n · area(Ki(d)) ≤ A, while the subfamily above covers na ≥ λA.2. The diameter of the union of all sets is at most 2d, by the assumption. Its area
is therefore at most πd2, by the well-known isoperimetric inequality. Thus λA ≤ a,and one single set will cover at least λA.
Comments 71 It is still unknown whether the second claim holds without any addi-tional assumption on the sets.
Problem 3.52. Let C be a regular polygon with k sides. Prove that for every n thereexists a planar set S(n) ⊂ R
2 such that any subset consisting of n points of S(n) canbe covered by C, but S(n) itself cannot be included in C.
Solution 3.52. Let r be the radius of the circle inscribed in C. For n ∈ Z+, let S(n)be the circle of radius r sec(π/2kn).
1.C cannot cover S(n). In fact, there is no disk of radius greater than r that can beincluded in C. Here is a proof for this somewhat obvious assertion. For a fixed radiusR, let X(R) be the set of centers of circles of radius R included in C. Then X(R) isconvex and invariant of rotations centered at the center O of C, of angle 2π/k. ThusX(R) is either empty or contains O. Thus R ≤ r .
2. Let P1, . . . , Pn ⊂ S(n) be arbitrary points of S(n). We show that C can coverP1, . . . , Pn. We fix a point Q in C at a distance of r sec π
2kn from O. We place Csuch that their centers coincide and we rotate C around its centerO such thatQ runsthrough S(n). While Q travels along the circle, some of the points Pi will be in theinterior of C and the others outside.
However, we see that Pi /∈ C if and only if Q ∈ ⋃nj=1Aj , where each Aj is an
arc of a circle from outside C, which is of length πkn
. More precisely, the arc Aj isdetermined by the inequalities
2πj
k≤ PiOQ ≤ 2
πj
k+ π
kn.
Therefore⋃nj=1Aj is a set of arcs of total length smaller than nkπ/kn · π and thus
S −⋃nj=1Aj has length π , and thus it is nonempty. Now rotate C untilQ lies inside
S − ⋃nj=1Aj . Then all the Pj belong to C, as claimed.
7.2 Combinatorial Geometry 169
Problem 3.53. Let M be a convex polygon and let S1, . . . , Sn be pairwise disjointdisks situated in the interior of M . Does there exist a partition M = D1 ∪ · · · ∪Dnsuch thatDi are convex disjoint polygons, each of which contains precisely one disk?
Solution 3.53. Let Oi, ri be the center and the radius of the disk Si . For a point X inthe plane, let us define hi(X) = XO2
i − r2i . We consider
Dk = {x ∈ M;hk(X) ≤ hi(X), for all i = 2, 3, . . . , n}.We will prove that the Di are convex and Si ⊂ Di . Fix k, for instance k = 1. Define
Hi = {X ∈ R2;hk(X) ≤ hi(X)}.
Then we claim that Hi are half-planes. Let i = 2 for simplicity. Choose the pointP ∈ |O1O2| such that |PO1|2 − r2
1 = |PO2|2 − r22 . Take for instance |PO1| =
|O1O2|2−(r21 −r2
2 )
2|O1O2| . If l is the line perpendicular to O1O2 at P and X ∈ l, then we have
|XO1|2 − r21 = |PO1|2 + |PX|2 − r2
1 = |PO2|2 + |PX|2 − r22 = |XO2|2 − r2
2 .If H2 is the half-plane that contains O1, then |XO1|2 − r2
1 ≤ |XO2|2 − r22 , for all
X ∈ H2, |XO1|2 − r21 ≥ |XO2|2 − r2
2 for allX /∈ H2. Therefore,H2 is the half-planeH2. Consequently,
Dk = M
n⋂
i=1,i �=kHn
is a convex set as an intersection of convex sets, and Sk ⊂ Dk . It is clear that⋃Di = M.
Problem 3.54. Consider an inscribable n-gon partitioned by means of n − 2 nonin-tersecting diagonals into n− 2 triangles. Prove that the sum of the radii of the circlesinscribed in these triangles does not depend on the particular partition.
Solution 3.54. If we have a triangle inscribed in a circle of centerO and radius R, ofinradius r , then the distances from O to the sides satisfy the well-known formula
∑
cyclic
d∗(O, side) = R + r.
Here d∗ is the signed distance, which is positive on one half-plane determined by theside and negative on the other half-plane. We write the corresponding formulas foreach triangle of the partition and sum up all the terms. Let d1, . . . , dn be the distancesfrom O to the sides of the n-gon. When looking at the signed distances from O toa diagonal, we observe that each signed distance is considered twice: once with theplus sign and once with the opposite sign. When these terms are summed, they willcancel. We then obtain
d1 + · · · + dn = (n− 2)R + r1 + · · · + rn,
and thus the sum of inradii r1 + r2 + · · · + rn does not depend on the partition.
170 7 Geometry Solutions
Problem 3.55. Prove that in an ellipse having semiaxes of lengths a and b and totallength L, we have L > π(a + b).
Solution 3.55. We cut the ellipse along the axes into four congruent pieces. We canfurther rearrange the four pieces by adding a square of side length b − a as in thefigure below:
We want to use the well-known Bonnesen’s isoperimetric inequality, which states that
L2 ≥ 4πA
for any planar convex domain of perimeter L and area A.The area of the new domain is A+ (b− a)2, where A was the area of the ellipse.
Moreover, one knows that the area of the ellipse isA = πab. Applying this inequalityto the domain pictured in the figure above, we obtain
L2 > 4π2ab + 4π(b − a)2 > π2(a + b)2,
whence the claim.
Problem 3.56. Let F be a convex planar domain and F ′ denote its image by a homo-thety of ratio − 1
2 . Is it true that one can translate F ′ in order for it to be contained inF ? Can the constant 1
2 be improved? Generalize to n dimensions.
Solution 3.56. Let ABC ⊂ F be a triangle of maximal area inscribed in F . Let Odenote the geometric centroid of ABC. We use the homothety of center O and ratio− 1
2 that transformsF intoF ′ andABC intoA′B ′C′. SinceO is the centroid ofABC,we find that A′ is the midpoint of the segment |BC|, B ′ is the midpoint of |CA|, andC′ is the midpoint of |AB|.
7.2 Combinatorial Geometry 171
We will show that F ′ ⊂ ABC. Assume the contrary, namely that there exists apoint X of F ′ that does not belong to the triangle ABC. By symmetry, it sufficesto consider that X belongs to the half-plane determined by the line BC, which doesnot contain A. Since BC and B ′C′ are parallel, it follows that the area of XB ′C′ isstrictly greater than the area of A′B ′C′. Let us consider the inverse image Y of X bythe homothety of center O and ratio − 1
2 . Then Y belongs to F and the area of YBCis strictly greater than the area of ABC, which contradicts our assumptions.
This proves that F ′ ⊂ ABC ⊂ F . The constant 12 is sharp in the case that F is a
triangle.The same proof shows that the image F ′ of a convex domain F in R
n by thehomothety of ratio − 1
ncan be translated in order to be contained in F . The constant
1n
is sharp, with equality for the simplex.
Problem 3.57. A classical theorem, due to Cauchy, states that a strictly convex poly-hedron in R
3 whose faces are rigid, must be globally rigid. Here, rigidity meanscontinuous rigidity, in the sense that any continuous deformation of the polyhedronin R
3 that keeps the lengths of edges fixed is the restriction of a deformation of rigidEuclidean motions of three-space. Prove that a 3-dimensional cube immersed in R
n
remains rigid for all n > 3.
Solution 3.57. Three sides are common to a vertex and determine a three-dimensionalspace R
3 ⊂ Rn. The 3-dimensional spaces determined by two adjacent vertices have
two sides in common, and therefore they coincide. Thus, the entire cube is immersedin a 3-dimensional subspace of R
n, and therefore it remains rigid.The same argument shows that a rigid polyhedron with the property that exactly
three sides meet at any vertex remains rigid after immersion in Rn, for all n > 3.
Problem 3.58. Consider finitely many great circles on a sphere such that not all ofthem pass through the same point. Show that there exists a point situated on exactlytwo circles. Deduce that if we have a set of n points in the plane, not all of them lyingon the same line, then there must exist one line passing through precisely two pointsof the given set.
Solution 3.58. Assume that the contrary holds. Consider the partition of the sphereinto (spherical) polygons, induced by the circles, and identify it with a polyhedron.Each vertex will have degree at least 6, since at least three great circles pass througheach point. If V is the number of vertices, E the number of edges, and F the numberof faces of the spherical polyhedron, then Euler’s formula states that
V − E + F = 2.
Moreover, by counting the edges in each face and summing over the faces, we obtain
2E = 3F3 + 4F4 + · · · ≥ 3F,
where Fk is the number of faces with k sides. Finally, counting the edges adjacent toeach vertex and summing over the vertices, we get
172 7 Geometry Solutions
2E = 6V6 + 7V7 + · · · ≥ 6V,
where Vk is the number of vertices of degree k, because Vk = 0 when k ≤ 5. Thus6(E + 2) = 6(V + F) ≤ 2E + 4E = 6E, which is absurd. Therefore, there exists avertex of valence smaller than 6 in which exactly two great circles intersect.
The second claim can be reformulated using projective duality: if we have n linesin the projective plane, not all passing through the same point, then there exists a pointlying on exactly two of the lines. This is equivalent to the problem considered above.
Comments 72 The problem was posed by Sylvester in 1839, and it was not settleduntil 1930, when T. Gallai gave a solution. The solution above is due to N. Steenrod.Moreover, it can be proved that the number of points where exactly two circles intersectis at least 3n/7; see for instance:
• H.S.M. Coxeter: The classification of zonohedra by means of projective diagrams,J. Math. Pures Appl. (9) 41 (1962), 137–156.
• G.D. Chakerian, Sylvester problem on collinear points and a relative,Amer. Math.Monthly 77 (1970), 164–167.
Problem 3.59. Given a finite set of points in the plane labeled with +1 or −1, andnot all of them collinear, show that there exists a line determined by two points in theset such that all points of the set lying on that line are of the same sign.
Solution 3.59. We use projective duality as above and reformulate the question asfollows: given n great circles on a sphere, to each one being assigned either +1 or−1, and not all passing through the same point, show that there exists a point lyingon at least two circles such that all circles containing that point have the same sign.
The circles determine a spherical polyhedron whose edges are labeled by +1 and−1. Assume that the result is not true. Then the edges around each vertex must haveboth the labels +1 and −1. By symmetry, the number of sign changes in the labelsof the edges that one meets when traveling-cyclically around each vertex is at least 4.
Denote by N the sum of all such sign changes over all vertices. Then N ≥ 4V .On the other hand, by counting the sign changes as we travel around the faces
and summing over all faces, we should again obtain N . Further, the number of signchanges in a k-sided face is even and at most k, so that
N ≤ 2F3 + 4F4 + 4F5 + 6F6 + 6F7 + · · · .Moreover, using Euler’s formula and
2E = 3F3 + 4F4 + · · ·we obtain
4V − 8 = 4E − 4F = 2(3F3 + 4F4 + 5F5 + · · · )− 4(F3 + F4 + · · · )= 2F3 + 4F4 + 6F5 + · · · ≥ N ≥ 4V,
which is a contradiction. Thus our claim follows.
7.2 Combinatorial Geometry 173
Comments 73 The result concerning the sign changes is valid more generally whensome edges are allowed to be unlabeled, and as such it is known as Cauchy’s lemma.This is an essential ingredient in proving the famous rigidity theorem of Cauchy, whichstates that a convex polyhedron in R
3 whose faces are rigid is globally rigid.
Problem 3.60. If Q is a given rectangle and ε > 0, then Q can be covered by theunion of a finite collection S of rectangles with sides parallel to those of Q in such away that the union of every nonoverlapping subcollection of S has area less than ε.
Solution 3.60. Observe first that every rectangle Q has a subset H that is the unionof a finite collection � of rectangles with parallel sides such that:
1. any two members of � overlap;2. area(H) > δ area(Q), where δ = exp(−1/ε);3. area(A) < ε
2 area(H), for any A ∈ �.
If Q = [0, a] × [0, b], we consider rectangles Ai that are parallel and have onevertex at (0, 0) and the other vertex on the curve of equation xy = δ area(Q). Thuseach rectangle has area δ. The curved triangle determined by the inequalities xy ≤δ area(Q), x, y ≥ 0, will be denoted by T . Moreover, we consider a collection of suchrectangles such that their union H satisfies area(H) > 1+ε
εarea(T ) = δ area(Q).
Further, area(A) = δ area(Q) < ε area(T ).Set Q0 = Q, H0 = H , �0 = � and assume that area(Q) = 1, for simplicity.
Then Q0 −H0 is a finite union of parallel rectangles Qi,0 and
area(Q0 −H0) < (1 − δ).
Choose in each rectangle Qi a set of type Hi,0 as above. The union of all such Hi,0and H0 will be denoted by H1, and the collection of rectangles will be denoted by�1. Then Q−H1 is also a finite union of parallel rectangles Qi,1 and
area(Q−H1) < (1 − δ)2 area(Q).
We continue this process by defining the setsHj and the collections�j of rectanglessuch that Q − Hj is a finite union of rectangles Qi,j whose complement has totalarea
area(Q−Hj) < (1 − δ)j area(Q).
We stop this process at the stage n, where n is large enough that (1 − δ)n area(Q) <ε/2. Consider then the union � of the collections �n obtained so far with the set ofrectangles in which Q−Hn decomposes.
If we consider a subfamily of disjoint rectangles from �, then there is at mostone element Ai from any collection �i,j covering Hi,j , because all elements in �i,joverlap each other. Thus the total area covered by these Ai is at most
∑
i,j
ε
2area(Hi,j ) <
ε
2.
Since the rectangles from Q−Hn contribute at most ε2 , we are done.
174 7 Geometry Solutions
Problem 3.61. Prove that the 3-dimensional ball cannot be partitioned into three setsof strictly smaller diameter.
Solution 3.61. Assume that we can divide the unit ball into three pieces of smallerdiameter and let X1, X2, X3 be the partition that it induces on the boundary sphere.Thus the diameter of each piece Xi is smaller than 1 − 10h, for some h > 0. Wechoose a very thin triangulation of the sphere with triangles that have edges smallerthan h. Denote byNj the union of those spherical triangles of the triangulation abovethat intersect the set Xj nontrivially. It follows that N1, N2, N3 are closed subsetsof the sphere that are domains, i.e., they are closures of open subsets. In particular,the boundary of each domain Nj is the union of several piecewise linear circles.Moreover, the union N1 ∪N2 ∪N3 covers the sphere.
Since each triangle has a small size, the diameter of Nj is at most 1 − 8h < 1.In particular, opposite points cannot belong simultaneously to the same set Nj . Letus denote by x∗ the point of the sphere opposite the point x. We know that the setN1 and its symmetric N∗
1 (which is the set of those points x∗ for which x ∈ N1)should be disjoint; otherwise, there would exist in N1 a pair of points at distance1. Consider the circles determined by the boundary circles of both N1 and N∗
1 . Thecircles associated to Nj are disjoint and distinct from the circles associated to N∗
1 ,because N1 ∩ N∗
1 = ∅. Thus, there is an even number of such circles. Observe thatk disjoint circles on the sphere divide the surface of the sphere into k + 1 connectedcomplementary regions. Thus, the collection of boundary circles associated to N1and N∗
1 will split the surface of the sphere into an odd number of connected regions,which we denote by R1, R2, . . . , R2m+1.
Consider the symmetric R∗j of the region Rj . There are two possibilities: either
R∗j is another connected regionRk that is disjoint fromRj , or elseR∗
j = Rj , meaningthat Rj is symmetric. The second case happens, for instance, when the region Rj isbounded by two circles, one coming fromN1 and the other fromN∗
1 . If there were nosymmetric regions Rj , then the total number of regions would be even, which wouldcontradict our previous claim. Thus there must be at least one connected symmetricregion Rj .
We consider a point x ∈ Rj and its symmetric x∗, which must also belong toRj , since the region is symmetric. If x ∈ N1, then all points of the sphere that areconnected to x by a path not hitting a boundary circle of N1 must also belong toN1. Thus, the connected component Rj will be contained in N1, and so N1 willcontain pairs of symmetric points, contradiction. This implies that x �∈ N1 and hencex ∈ N2 ∪ N3. Let us assume that x ∈ N2. Then its symmetric x∗ is not in N2, sinceN2 has diameter smaller than 1 and thus x∗ ∈ N3. Further, Rj is connected and thusthere exists a path γ within Rj that connects x to x∗. One knows that N2 and N3 aredomains covering the path γ . Moreover, there exists a point z ∈ γ that belongs toboth N2 and N3. In fact, consider the point z ∈ γ closest to x∗ and that belongs toN2. There exists such a point sinceN2 is closed. If z = x∗, then both x and x∗ belongto N2, contradiction. If z �= x∗, then the points on the right of z should belong to N3and thus z belongs to the closure of N3 and thus to N3.
7.3 Geometric Inequalities 175
Now z ∈ Rj and thus its symmetric z∗ also belongs to Rj . This means thatz, z∗ �∈ N1. But z∗ can belong neither to N2, since N2 does not contain oppositepoints, nor to N3, for the same reason. This contradiction proves the claim.
Comments 74 More generally, it is known that the unit n-ball (or equivalently, the(n−1)-sphere in R
n) cannot be partitioned into n pieces of strictly smaller diameter.This is related to Borsuk’s problem discussed in comments to Problem 3.38. The resultwas proved by Lusternik and Shnirelman in 1930 and, independently, by Borsuk in1932, but the higher-dimensional case n ≥ 4 requires deeper methods from topology,and this opened a new chapter in algebraic topology. The interested reader mightconsult the survey of Steinlein, which presents both historical remarks and a largenumber of references.
• K. Borsuk: Über die Zerlegung einer n-dimensionalen Vollkugel in n-Mengen,Verh. International Math. Kongress Zürich 1932, 192.
• L.A. Lusternik and L.G. Shnirelman: Topological Methods for Variational Prob-lems, Moscow, 1930.
• H. Steinlein: Borsuk’s antipodal theorem and its generalizations and applications:a survey, Topological Methods in Nonlinear Analysis, 166–235, Sém. Math. Sup.,95, Presses Univ. Montreal, Montreal, 1985.
• H. Steinlein: Spheres and symmetry: Borsuk’s antipodal theorem. Topol. MethodsNonlinear Anal. 1 (1993), 1, 15–33.
7.3 Geometric Inequalities
Problem 3.62. If a, b, c, r, R are the usual notations in the triangle, show that
1
2rR≤ 1
3
(∑ 1
a
)2
≤∑ 1
a2 ≤ 1
4r2 .
Solution 3.62. We derive from∑( 1a
− 1b)2 ≥ 0 that
∑ 1
bc≤ 1
3
(∑ 1
a
)2
≤∑ 1
a2 .
Moreover, the usual Cauchy–Schwarz inequality shows that (p − a)(p − b) ≤ c2
4 ,and its analogues. Then we have
1
2rR= 2p
abc= a + b + c
abc= 1
ab+ 1
bc+ 1
ca≤ 1
3
(∑ 1
a
)2
≤∑ 1
a2
≤ 1
4
(∑ 1
(p − b)(p − c)
)= p
4(p − a)(p − b)(p − c)= 1
4r2 .
Problem 3.63. If a, b, c are the sides of a triangle, then prove that (b+c)2
4bc ≤ mawa
andb2+c2
2bc ≤ maka
, wherema,wa, ka denotes the respective lengths of the median, bisector,and altitude issued from A.
176 7 Geometry Solutions
Solution 3.63. We have the following formulas computing ma,wa , and ka :
ma = 1
2
√2(b2 + c2
) − a2, wa =√bc
b + c
√(b + c)2 − a2,
ka = bc
b2 + c2
√2(b2 + c2
) − a2.
From the triangle inequality, we have a2 > (b− c)2, which implies furthermore that
(b + c)2 − a2 < 4bc(b − c)2
4bc≤ (b − c)2
(b + c)2 − a2 ,
with equality only if b = c. Adding 1 in both members yields
(b + c)2
4bc≤ 2b2 + 2c2 − a2
(b + c)2 − a2 ; henceb + c
2√bc
≤√
2b2 + 2c2 − a2
(b + c)2 − a2 .
This leads to
(b + c)2
4bc≤ b + c
2√bc
√2(b2 + c2)− a2
(b + c)2− a2 = ma
wa.
For the second inequality, it suffices to observe that
ma
ha≥ ma
ka= b2 + c2
2bc.
Notice that equality in the first case implies b = c, while for the second case eitherb = c or a2 = b2 + c2.
Problem 3.64. If S(x, y, z) is the area of a triangle with sides x, y, z, prove that√S(a, b, c)+ √
S(a′, b′, c′) ≤ √S(a + a′, b + b′, c + c′).
Solution 3.64. We make the following change of variables: s = (a + b + c)/2,t = s− a, u = s− b, and v = s− c. Using Heron’s formula, the inequality becomes
4√stuv + 4
√s′t ′u′v′ ≤ 4
√(s + s′)(t + t ′)(u+ u′)(v + v′).
We prove first that for positive x, x′, y, y′, we have
√xy + √
x′y′ ≤ √(x + x′)(y + y′).
In fact, this follows from xy + x′y′ + 2√x′y · y′x ≤ xy + x′y′ + xy′ + yx′, and
equality holds when the geometric means of x′y and y′x are equal.By applying the last inequality to x = √
st and y = √uv twice, we obtain the
claim. Moreover, the equality is attained only when t/t ′ = u/u′ = v/v′ = s/s′,which amounts to a/a′ = b/b′ = c/c′.
7.3 Geometric Inequalities 177
Problem 3.65. It is known that in any triangle we have the inequalities
3√
3r ≤ p ≤ 2R + (3√
3 − 4)r,
where p denotes the semiperimeter. Prove that in an obtuse triangle, we have
(3 + 2√
2)r < p < 2R + r.
Solution 3.65. We havea + b
c= cos A−B
2
sin C2
.
Since C > π2 , we have sin C
2 >√
22 and thus c
√2 > a + b. Therefore
2p√
2 > (a + b)(√
2 + 1) ⇒ 8p2
> (3 + 2√
2)(a + b)2 > 4(3 + 2√
2)ab
> 8(3 + 2√
2)S,
where S is the area. This implies that
p >
(3 + 2
√2)S
p= (
3 + 2√
2)r.
Finally, recall that r = (p− c) tan C2 > (p− c) and so p < c+ r = 2R sinC + r <
2R + r .
Problem 3.66. Prove the Euler inequality
R ≥ 2r.
Solution 3.66.|OI |2 = R2 − 2Rr.
Problem 3.67. Prove that in a triangle we have the inequalities
36r2 ≤ a2 + b2 + c2 ≤ 9R2.
Solution 3.67. We have
p
3= 1
3(p − a + p − b + p − c) ≥ 3
√(p − a)(p − b)(p − c) = 3
√S2
p= 3
√r2p,
and hence p2 ≥ 27r2. Thus
a2 + b2 + c2 ≥ 4
3p2 ≥ 36r2.
The second inequality is a consequence of the following identity:
|OH |2 = 9R2 − (a2 + b2 + c2),
where O is the circumcenter and H the orthocenter.
178 7 Geometry Solutions
Problem 3.68. 1. Let ABC and A′B ′C′ be two triangles. Prove that
a2
a′ + b2
b′ + c2
c′≤ R2 (a
′ + b′ + c′)2
a′b′c′.
2. Derive that
a2 + b2 + c2 ≤ 9R2,
cosA cosB cosC ≤ 1
8.
Solution 3.68. 1. The inequality from the statement is equivalent to
a′2 + b′2 + c′2 − 2(b′c′ cosA′′ + a′c′ cosB ′′ + b′a′ cosC′′) ≥ 0,
where A′′ = π − 2A,B ′′ = π − 2B,C′′ = π − 2C. Moreover, this can be written as
(a′ − b′ cosC′′ − c′ cosB ′′)2 + (b′ sinC′′ − c′ sinB ′′)2 ≥ 0.
A necessary condition for equality is the vanishing of the second square, i.e.,
a′
sinA′′ = b′
sinB ′′ = c′
sinC′′ .
But A′′ ∈ (−π, π); therefore A′′, B, C′′ are the angles of a triangle that is similar toA′B ′C′. Therefore, we obtain equality if and only if ABC is an acute triangle andA′B ′C′ is similar to the orthic triangle of the triangle ABC.
2. If A′B ′C′ is equilateral, then a2 + b2 + c2 ≤ 9R2, which is equivalent tosin2 A+ sin2 B+ sin2 C ≤ 9
4 . Observe that∑
sin2 A = 2 +2 cosA cosB cosC, andtherefore we obtain cosA cosB cosC ≤ 1
8 .
Problem 3.69. Prove that the following inequalities hold in a triangle:
4∑
cyclic
hAhB ≤ 12S√
3 ≤ 54Rr ≤ 3∑
cyclic
ab ≤ 4∑
cyclic
rArB.
Solution 3.69. 1. We have
4∑
cyclic
rArB =⎛
⎝∑
cyclic
a
⎞
⎠
2
≥ 3∑
cyclic
ab.
2. In order to prove that18Rr ≤
∑
cyclic
ab,
we start from the identity1
ha+ 1
hb+ 1
hc= 1
r.
7.3 Geometric Inequalities 179
Thus
2abc
4R
(1
a+ 1
b+ 1
c
)= 2S
(1
a+ 1
b+ 1
c
)= ha + hb + hc ≥ 9
∑ 1ha
≥ 9r,
leading to the inequality we wanted.3. Recall that Jensen’s inequality states that if f is a concave (i.e., f ′′(x) ≤ 0)
smooth real function on some interval J , then
λ1f (x1)+ λ2f (x2)+ · · · + λnf (xn) ≤ f (λ1x1 + λ2x2 + · · · + λn + xn)
for all xj ∈ J and λi ≥ 0 such that λ1 + λ2 + · · · + λn = 1.Notice now that the function cos is concave on
[0, π2
], and by Jensen’s inequality
we have
cosA
2+ cos
B
2+ cos
C
2≤ 3
√3
2.
Therefore, by the means inequality,
cosA
2cos
B
2cos
C
2≤
(cos A2 + cos B2 + cos C2
3
)3
≤ 3√
3
8.
We have then
S = abc
4R= 4rR cos
A
2cos
B
2cos
C
2≤ 3
√3
2rR.
4. We have
∑
cyclic
hAhB = hAhBhC
(1
ha+ 1
hb+ 1
hc
)= 2S2
R
1
r≤ 3
√3S.
Problem 3.70. Prove that in an any triangle ABC, we have
√1 + 8 cos2 B
sinA+
√1 + 8 cos2 C
sinB+
√1 + 8 cos2 A
sinC≥ 6.
Solution 3.70. By applying the means inequality, the left-hand side is greater than
3
√√√√√3
⎛
⎝∏
cyclic
(√
1 + 8 cos2 A
⎞
⎠∏
cyclic
3
√1
sinA,
and it is sufficient to prove that
F(A,B,C) =∏
cyclic
1 + 8 cos2 A
sin2 A≥ 64.
180 7 Geometry Solutions
Let us find the extremal points of F , that satisfy
∂F
∂A= ∂F
∂B= ∂F
∂C.
We have1
F· ∂F∂A
= − 18 cotA
1 + 8 cos2 A,
and thus we have
sinB cosC(1 + 8 cos2 B) = sinC cosB(1 + 8 cos2 C).
This implies that sin(B − C)+ 4 cosB cosC(sin 2B − sin 2C) = 0 and hence
sin(B − C)(1 − 8 cosA cosB cosC) = 0.
We claim that this happens only if A = B = C. In fact, if the triangle is obtuse, then
cosA cosB cosC ≤ 0 <1
8
and hence B = C = A.Otherwise, cos is positive and
cosA+ cosB + cosC = 1 + 4 sinA
2sin
B
2sin
C
2= 1 + 4
∏
cyclic
√(p − b)(p − c)bc
= 1 + r
R≤ 3
2,
and so by the means inequality,
cosA cosB cosC ≤(
cosA+ cosB + cosC
3
)3
≤ 1
8.
Equality holds above only if the triangle is equilateral.This shows that the only extremal points are A = B = C = π
3 , and one verifiesthat it corresponds to a minimum, so that F(A,B,C) ≥ 64, as claimed.
Problem 3.71. Let P be a point in the interior of the triangle ABC. We denote byRa,Rb, Rc the distances from P toA,B,C and by ra, rb, rc the distances to the sidesBC,CA,AB. Prove that
∑
cyclic
R2a sin2 A ≤ 3
∑
cyclic
r2a ,
with equality if and only if P is the Lemoine point (i.e., the symmedian point).
7.3 Geometric Inequalities 181
Solution 3.71. Let G be the centroid of the poder triangle DEF of P , obtained byprojecting P to the sides of ABC. We have then
3∑
cyclic
r2a = 3(|DG|2 + |EG|2 + |FG|2 + 3|PG|2) ≥ 3(|DG|2 + |EG|2 + |FG|2)
= |DE|2 + |EF |2 + |FD|2 =∑
cyclic
R2a sin2 A.
We obtain equality when P is the centroid of its poder triangle. Therefore P is thesymmedian point.
Comments 75 We can generalize this type of inequality. We have, for instance,
xR21 + yR2
1 + zR21 ≥ a2yz+ b2xz+ c2xy
x + y + z
for every x, y, z ∈ R such that x+ y+ z > 0. The equality holds above if and only if
x
F1= y
F2= z
F3,
where F1, F2, F3 denote the areas of BPC, APC, and APB, respectively.
Comments 76 With every inequality �(Ra,Rb, Rc, a, b, c) ≥ 0, valid for any tri-angle of sides a, b, c, we can associate a dual inequality in a new set of variables,�(ra, rb, rc, Ra sinA,Rb sinB,Rc sinC) ≥ 0, obtained by replacing the originaltriangle by the poder triangle associated to P . Then the distances from P to thevertices of the poder triangle are ra, rb, rc and the sides of the poder triangle areRa sinA,Rb sinB,Rc sinC.
For instance, the dual of the inequality from the previous remark is
xr2a + yr2
b + zr2c ≥ yzR2
a sin2 A+ xzR2b sin2 B + xyR2
c sin2 C
x + y + z.
If x = y = z, then this is the inequality of our problem. Applying the inequality fromthe previous remark to the right-hand side, we also obtain
∑yzR2
1 sin2 A
x + y + z≥ 4S2
a2
x+ b2
y+ c2
z
.
Problem 3.72. Prove the inequalities
16Rr − 5r2 ≤ p2 ≤ 4R2 + 4Rr + 3r2.
Solution 3.72. Let I be the incenter of the triangle and H be the orthocenter. Directcalculations show that
|IH |2 = 2r2 − 4R2 cosA cosB cosC,
182 7 Geometry Solutions
and replacing the term
4R2 cosA cosB cosC = p2 − (2R + r)2,
we find that|IH |2 = 4R2 + 4Rr + 3r2 − p2.
The next step is to use Euler’s theorem, which states that O,G, and H are collinearand |OG|
|GH | = 1
2,
where G denotes the centroid of the triangle. We compute the length of |GI | usingthe triangle HIO as follows:
|GI |2 = 2
3|IO|2 + 1
3|IH |2 + 2
9|OH |2.
We substitute
|OI |2 = R2 − 2Rr, |OH |2 = 9R2 − (a2 + b2 + c2)
and derive that
|GI |2 = r2 − 1
3p2 + 2
9
(a2 + b2 + c2).
Moreover, 4p2 = 2(a2 + b2 + c2
) + 16Rr + 4r2, and so
p2 − 16Rr + 5r2 = 8|GI |2.Thus the inequality follows.
Comments 77 Using 2p2 = 2r(4R+ r)+ a2 + b2 + c2 and S = rp, we derive alsothe inequalities
12r(2R − r) ≤ a2 + b2 + c2 ≤ 4r2 + 8R2,
r3(16R − 5r) ≤ S2 ≤ r2(3r2 + 4rR + 4R2).
• O. Bottema, R.Ž. Djordjevic, R.R. Janic, D.S. Mitrinovic, and P.M. Vasic: Geo-metric Inequalities, Wolters–Noordhoff Publishing, Gröningen 1969.
Problem 3.73. Prove the following inequalities, due to Roché:
2R2 + 10Rr − r2 − 2(R − 2r)√R2 − 2Rr
≤ p2 ≤ 2R2 + 10Rr − r2 + 2(R − 2r)√R2 − 2Rr.
Solution 3.73. Suppose that a ≥ b ≥ c. By direct calculation, we compute the areaof this triangle as follows:
area(HIO) = 2R2 sin
(B − C
2
)sin
(A− C
2
)sin
(A− B
2
)= (b − c)(a − c)(a − b)
8r.
7.3 Geometric Inequalities 183
Expressing R and r in terms of a, b, c, we can conclude with the identity
(area(HIO))2 = −p4 + 2(2R2 + 10Rr − r2)p2 − r(4R + r)3.
Since this binomial in p2 is positive, the value of p2 lies between the two real rootsof the polynomial, which yields the claim.
Comments 78 W.J. Blundon gave another proof, using the following identity:
−p4 + 2(2R2 + 10Rr − r2)p2 − r(4R + r)3 = 1
4r2 (a − b)2(b − c)2(c − a)2.
• D.S. Mitrinovic, J.E. Pecaric, V. Volenec: Recent Advances in Geometric Inequal-ities, Math. Appl. (East European Series), 28, Kluwer Academic Publ., Dordrecht,1989.
8
Analysis Solutions
Problem 4.1. Prove that z ∈ C satisfies |z| − �z ≤ 12 if and only if z = ac, where
|c − a| ≤ 1. We denote by �z the real part of the complex number z.
Solution 4.1. We have the identity
|ac| − � ac = 1
2|c − a|2 − 1
2(|c| − |a|)2,
and the “if” part follows. For the converse, we consider a, c such that |a| = |c| =|z|1/2.
Problem 4.2. Let a, b, c ∈ R be such that a+2b+3c ≥ 14. Prove that a2+b2+c2 ≥14.
Solution 4.2. From (a − 1)2 + (b − 2)2 + (c − 3)2 ≥ 0, we obtain a2 + b2 + c2 ≥2(a + 2b + 3c)− 14 ≥ 14.
Comments 79 More generally, if wi ∈ R+ and ai ∈ R are such that∑ni=1 aiwi ≥∑n
i=1w2i , then
∑ni=1 a
2i ≥ ∑n
i=1w2i .
Problem 4.3. Let fn(x) denote the Fibonacci polynomial, which is defined by
f1 = 1, f2 = x, fn = xfn−1 + fn−2.
Prove that the inequality
f 2n ≤ (
x2 + 1)2(x2 + 2
)n−3
holds for every real x and n ≥ 3.
Solution 4.3. Since f3(x) = x2 + 1, the inequality is trivially satisfied for n = 3. Weproceed by induction on n:
186 8 Analysis Solutions
f 2n+1(x) = [xfn(x)+ fn−1(x)]
2
≤ [x(x2 + 1)(x2 + 2)(n−3)/2 + (x2 + 1)(x2 + 2)(n−4)/2]2
≤ (x2 + 1)2(x2 + 2)n−2[x(x2 + 2)−1/2 + (x2 + 2)−1]2
< (x2 + 1)2(x2 + 2)k−2.
We have used above that x(x2 + 2
)−1/2 + (x2 + 2
)−1< 1, which is a consequence
of x4 + 2x2 <(x2 + 1
)2.Second proof. One proves by induction that
fn(x) = det
⎛
⎜⎜⎜⎜⎜⎝
x −1 0 0 · · · 0 01 x −1 0 · · · 0 00 1 x −1 · · · 0 0......
......
......
0 0 0 0 · · · 1 x
⎞
⎟⎟⎟⎟⎟⎠
⎫⎪⎪⎪⎪⎪⎬
⎪⎪⎪⎪⎪⎭
n− 1.
Let us recall now the Hadamard inequality, which gives an upper bound for thedeterminant of an arbitrary k × k matrix, as follows:
(det(aij ))2 ≤
k∏
j=1
(k∑
i=1
a2ij
)
.
This yields the claimed inequality.
Problem 4.4. Prove the inequality
min((b − c)2, (c − a)2, (a − b)2
) ≤ 1
2
(a2 + b2 + c2).
Generalize to min1≤k<i≤n(ak − ai)2.
Solution 4.4. We will prove that for any real numbers ai , we have
mini<j
(ai − aj )2 ≤ μ2(a2
1 + · · · + a2n
),where μ2 = 12
n(n2 − 1).
There is no loss of generality if we suppose that a1 ≤ · · · ≤ an and∑ni=1 a
2i = 1.
Then∑
1≤i<j≤n(ai − aj )
2 = n
n∑
i=1
a2i −
(n∑
i=1
ai
)2
.
If ai+1 − ai > μ > 0 for all i ≤ n− 1, then (aj − ai)2 > (i − j)2μ2 and hence
∑(ai − aj )
2 > μ2∑
1≤i<j≤n(i − j)2 = μ2 n
2(n2 − 1
)
12= n.
8 Analysis Solutions 187
This implies that
n < n
n∑
i=1
a2i − ( n∑
i=1
ai)2 ≤ n
n∑
i=1
a2i = n,
which is absurd. Therefore, there exist i �= j such that (ai − aj )2 ≤ μ2.
Problem 4.5. Let a1, . . . , an be the lengths of the sides of a polygon andP its perime-ter. Then
a1
P − a1+ a2
P − a2+ · · · + an
P − an≥ n
n− 1.
Solution 4.5. Set P − ai = yi . Then∑ni=1 yi = nP − ∑n
i=1 ai = (n − 1)P . Now
we express ai = P − yi =(
1n−1
∑nj=1 yj
)− yi = 1
n−1
(∑j �=i yj − (n− 2)yi
).
Thus we can write
n∑
i=1
ai
S − ai= 1
n− 1
n∑
j=1
∑
j �=i
(yj
yi− n− 2
n− 1
).
Further,
n∑
j=1
∑
j �=i
(yj
yi− n− 2
n− 1
)=
n∑
j=1
∑
i<j
(yj
yi+ yi
yj− 2
n− 2
n− 1
)≥ 2n
n− 1,
where we have regrouped the terms with indices i, j together with those with indicesj, i and used the fact that
yjyi
+ yiyj
≥ 2.
Problem 4.6. Assume that a, b, and c are positive numbers that cannot be the sidesof a triangle. Then the following inequality holds
(abc)2(a + b + c)2(a + b − c)(a − b + c)(b − a + c)
≥ (a2 + b2 + c2)3(
a2 + b2 − c2)(a2 − b2 + c2)(b2 − a2 + c2).
Solution 4.6. Let us note the following identity:
(a + b + c)(a − b + c)(a + c − b)(−a + b + c) = −∑
a4 + 2∑
a2b2,
where all sums are cyclic sums over the three letters involved. One multiplies bothmembers of the identity by
∑a2 and obtains
(a2 + b2 − c2)(a + b + c)(a − b + c)(−a + b + c)(a + b − c)
= −∑
a6 +∑
a4b2 + 6a2b2c2
= (−a2 + b2 + c2)(a2 − b2 + c2)(a2 + b2 − c2).
This help us to write the inequality above in the following form:
188 8 Analysis Solutions
2p(2p − a)(2p − b)(2p − c)((abc)2(2p)2 − (
a2 + b2 + c2)4)
+ 8(abc)2(a2 + b2 + c2)3 ≥ 0,
where 2p = a + b+ c. Let us note that (abc)2(a + b+ c)2 − (a2 + b2 + c2
)4 ≤ 0.In fact,
abc(a + b + c) = a2bc + b2ca + c2ab ≤ 1
2
∑a2(b2 + c2)
= 1
2
((∑a2
)2
−∑
a4
)
≤ (a2 + b2 + c2)2
.
Finally, if a, b, c do not determine a triangle, then
2p(2p − a)(2p − b)(2p − c) ≤ 0,
and our inequality follows.
Problem 4.7. Prove that sin2 x < sin x2, for 0 < x ≤ √π2 .
Solution 4.7. For 1 ≤ x2 ≤ π2 , the inequality is obvious. For 0 < x ≤ 1, we have
0 < t2 < t < 1 and hence 0 < cos t < cos t2. In particular, we have 2 sin t cos t <2t cos t2. Integrating from 0 to x, we obtain the claimed inequality.
Problem 4.8. Let P(x) = a0 + a1x + · · · + anxn be a real polynomial of degree
n ≥ 2 such that
0 < a0 < −[n2
]∑
j=1
1
2k + 1a2k.
Show that P(x) has a real zero in (−1, 1).
Solution 4.8. The hypothesis implies that
∫ 1
−1P(x) dx =
[n2
]∑
j=0
1
2k + 1a2k < 0.
Thus there exists x ∈ (−1, 1) with P(x) < 0. Since P(0) = a0 > 0, we can applythe intermediate value theorem to find that P has a real zero in (−1, 1).
Problem 4.9. Find the minimum of β and the maximum of α for which
(1 + 1
n
)n+α≤ e ≤
(1 + 1
n
)n+β
holds for all n ∈ Z+.
8 Analysis Solutions 189
Solution 4.9. Considering the logarithms, we have
αmax = infn
{1
log(1 + 1/n)− n
},
βmin = supn
{1
log(1 + 1/n)− n
}.
The function F(x) = 1log(1+1/x) − x is monotonically increasing for positive x, since
the derivative F ′(x) is positive. Therefore
αmax = 1
log 2− 1
andβmin = lim
n→∞F(n).
Using the Maclaurin series for the function log(1 + 1/x), we obtain
F(n) = 11n
− 12n2 + O( 1
n3 )− n,
whence limn→∞ F(n) = 12 = βmin.
Problem 4.10. Prove that for nonnegative x, y, and z such that x + y + z = 1, thefollowing inequality holds:
0 ≤ xy + yz+ zx − 2xyz ≤ 7
27.
Solution 4.10. We have x, y, z ∈ [0, 1] and thus 0 ≤ xyz ≤ 1, which turns intoxy + yz+ zx − 2xyz ≥ 3 3
√x2y2z2 − 2xyz ≥ 3xyz− 2xyz = xyz ≥ 0. This yields
the first inequality.Also, if one of the three variables x, y, z is at least 1
2 , then (1 − 2x)(1 − 2y)(1 −2z) ≤ 0 < 1
27 . Observe that it is not possible for two among the three variablesto be greater than 1
2 , because this would imply that the third is negative. Further, ifx, y, z ≤ 1
2 , then, by the means inequality, (1 − 2x)(1 − 2y)(1 − 2z) ≤ 127 . This
implies that
1 − 2(x + y + z)+ 4(xy + yz+ zx)− 8xyz ≤ 1
27
and thus xy + yz+ zx − 2xyz ≤ 727 .
Second Solution:Let f (x, y, z) = xy + yz+ zx − 2xyz and the associated Lagrange multiplier
F(x, y, z, λ) = f − λ(x + y + z− 1).
190 8 Analysis Solutions
The critical points of F are given by the formula
∂F
∂x= ∂F
∂y= ∂F
∂z= ∂F
∂λ= 0.
Therefore, we obtain the system of equations
x + y − 2xy − λ = 0,
x + z− 2xz− λ = 0,
y + z− 2yz− λ = 0,
x + y + z− 1 = 0.
If one variable, say z, is not equal to 12 , then x = z−λ
1−2z = y. If x = y �= 12 , then
x = z = y−λ1−2y . Thus x = y = z = 1
3 .
If x = y = 12 , then z = 0. Therefore, we have the solutions
( 13 ,
13 ,
13
),( 1
2 ,12 , 0
)
and their circular permutations.The Jacobian matrix reads
J (x, y, z) = det
⎛
⎜⎜⎝
∂2F∂x2
∂2F∂x∂y
∂2F∂x∂z
∂2F∂y∂x
∂2F∂y2
∂2F∂y∂z
∂2F∂z∂x
∂2F∂z∂y
∂2F∂z2
⎞
⎟⎟⎠ = det
⎛
⎝0 1 − 2z 1 − 2x
1 − 2z 0 1 − 2y1 − 2y 1 − 2x 0
⎞
⎠ ,
and one computes
J
(1
3,
1
3,
1
3
)= 2
27> 0, J
(1
2,
1
2, 0
)= 0.
Now f( 1
2 ,12 , 0
) = 14 < f
( 13 ,
13 ,
13
) = 727 , and so the only extremal point is a
maximum at( 1
3 ,13 ,
13
).
Problem 4.11. Consider the sequence of nonzero complex numbers a1, . . . , an, . . .
with the property that |ar − as | > 1 for r �= s. Prove that
∞∑
n=1
1
a3n
converges.
Solution 4.11. Let Sk = {n ∈ Z+; k < |an| ≤ k+1}. The disksDn = {z; |z−an| ≤
1/2}
are pairwise disjoint, according to the hypothesis. If one considers those disksDn for which n ∈ Sk , then these disks are contained in the annulus
Ck ={z ∈ C; k − 1
2≤ |z| ≤ k + 3
2
}.
Thus the total area covered by these disks is bounded from above by the area of theannulus, i.e.,
8 Analysis Solutions 191
card(Sk)π
4≤ π
((k + 3
2
)2
−(k − 1
2
)2)
= 2π(k + 1),
and thus card(Sk) ≤ 8(2k + 1) if k > 0. In the same way, for k = 0, we obtaincard(S0) ≤ 9:
∑
n∈Sk
1
|an|3 ≤ card(Sk)
k3 ≤ 8(2k + 1)
k3 ≤ 24
k2 ,
∞∑
n=1
1
|an|3 =∞∑
k=0
∑
n∈Sk
1
|an|3 ≤∑
n∈S0
1
|an|3 +∞∑
k=1
24
k2 < ∞,
and the series converges absolutely.
Problem 4.12. Consider the sequence Sn given by
Sn = n+ 1
2n+1
n∑
i=1
2i
i.
Find limn→∞ Sn.
Solution 4.12. We have
Sn+1 = n+ 2
2n+2
n+1∑
i=1
2i
i= n+ 2
2(n+ 1)(Sn + 1),
so that
Sn+2 − Sn+1 = (n+ 2)2(Sn+1 − Sn)− Sn+1 − 1
2(n+ 1)(n+ 2).
Since Sn ≥ 0, we obtain Sn+2−Sn+1 ≤ 0, and the sequence Sn is bounded decreasing,thus convergent. If we set s for its limit, then s satisfies
s = limn→∞
n+ 2
2(n+ 1)(Sn + 1) = 1
2(s + 1)
and thus s = 1.
Comments 80 By integrating the geometric series formula, we obtain
limn→∞
n
an
n∑
i=1
ai−1
i= 1
a − 1.
For a = 2, we recover the previous problem.
Problem 4.13. Prove that whenever a, b > 0, we have∫ 1
0
ta−1
1 + tbdt = 1
a− 1
a + b+ 1
a + 2b− 1
a + 3b+ · · · .
192 8 Analysis Solutions
Solution 4.13. Consider the identity
ta−1
1 + tb= ta−1
(n−1∑
k=0
(−1)ktkb + (−1)ntbn
1 + tb
)
.
Integrating between 0 and 1 and using the fact that
limn→∞
∣∣∣∣
∫ 1
0
tbn
1 + tbdt
∣∣∣∣ = 0,
we obtain the equality from the statement.
Problem 4.14. Let a, b, c, d ∈ Z∗+, and r = 1 − a
b− c
d. If r > 0 and a + c ≤ 1982,
then r > 119833 .
Solution 4.14. We have three cases to analyze:1. If b, d ≥ 1983, then r ≥ 1 − a+c
1983 ≥ 1 − 19821983 >
119833 .
2. If b, d ≤ 1983, then r = bd−ad−bcbd
> 0, so that bd − ad − bc ≥ 1. Thusr ≥ 1/bd > 1/19833.
3. Suppose now that b < 1983 < d. If d > 19832 and r < 119833 , then c
d<
198219832 <
11983 ; thus 1− a
b< 1
1983 . This implies that b > 1983(b−a) > 1983 because
a ≤ b + 1, which is absurd. If d ≤ 19832, then, as in the case 2 above, we haver ≥ 1
bd> 1
19833 .
Problem 4.15. Let ai ∈ R. Prove that
nmin(ai) ≤n∑
i=1
ai − S ≤n∑
i=1
ai + S ≤ nmax(ai),
where (n− 1)S2 = ∑1≤i<j≤n(ai − aj )
2, with S �= 0, and with equality if and onlyif a1 = · · · = an.
Solution 4.15. Let us assume a1 ≤ a2 ≤ a3 ≤ · · · ≤ an. Then
S2 = 1
n− 1
n∑
i=2
i−1∑
j=1
(ai − aj )2
≤ 1
n− 1
n∑
i=2
(i − 1)(ai − a1)2 ≤
n∑
i=2
(ai − a1)2 ≤
(n∑
i=1
(ai − a1)
)2
.
Extracting the radicals, we obtain na1 ≤ ∑ni=1 ai − S. Similarly,
S2 ≤ 1
n− 1
n−1∑
j=1
(n− j)(an − aj )2 ≤
⎛
⎝n−1∑
j=1
(an − aj )
⎞
⎠
2
,
and thus∑nk=1 ak + S ≤ nan. It is now clear that equality above implies S = 0, and
therefore a1 = a2 = · · · = an.
8 Analysis Solutions 193
Problem 4.16. Consider a periodic function f : R → R, of period 1 that is nonneg-ative, concave on (0, 1) and continuous at 0. Prove that for all real numbers x andnatural numbers n the following inequality f (nx) ≤ nf (x) is satisfied.
Solution 4.16. It is sufficient to discuss the case 0 < x < 1. Since f is concave, wehave f (ξ)
ξ<
f (x)x
for any ξ with x < ξ < 1. Thus f (ξ) < ξxf (x). If 0 < ξ < x,
then similarly f (ξ) < 1−ξ1−x f (x). Let ξ = {nx}, where the brackets {a} denote the
fractional part a − [a].If ξ > x, then we apply the first inequality above, and use the periodicity and the
fact that ξx
≤ n in order to get the claim.If ξ ≤ x, then we have two cases:1. If x ≤ 1 − 1
n, then 1−ξ
1−x ≤ 11/n = n, and we apply the second inequality above.
2. If x = 1− 1n+α for some 0 < α < 1
n, then nx = n−1+nα, and so {nx} = nα.
Since 1−ξ1−x = 1−nα
1n−α = n, we use the second inequality again and the claim follows.
Comments 81 Whenf (x) = | sin x|, we obtain the well-known inequality | sin nx| ≤n| sin x|.Problem 4.17. Let (an) be a sequence of positive numbers such that
limn→∞
(a1 + · · · + an)
n< ∞, lim
n→∞an
n= 0.
Does this imply that limn→∞a2
1+···+a2n
n2 = 0?
Solution 4.17. The answer is yes. If the sequence an is bounded, it is obvious. If thesequence is not bounded, consider the subsequence ank defined by the properties thatn1 is the smallest integer such that an1 > a1, and nk the smallest integer nk > nk−1such that ank > ank−1 .
If nk ≤ n < nk+1, then we have
0 ≤ a21 + · · · + a2
n
n2 <
(a1 + · · · + an
n
)ank
nk.
Passing to the limit, we find that the claim holds.
Problem 4.18. Show that for a fixed m ≥ 2, the following series converges for asingle value of x:
S(x) = 1 + ...+ 1
m− 1− x
m+ 1
m+ 1+ · · · + 1
2m− 1− x
2m
+ 1
2m+ 1+ · · · + 1
3m− 1− x
3m+ · · · .
Solution 4.18. Let us denote by Sn(x) = ∑nk=1(
1(k−1)m+1 + · · · + 1
km−1 − xkm) the
truncation of S(x). Assume that S(x) and S(y) are finite for two distinct values xand y.
194 8 Analysis Solutions
We now observe that Sn(x)− Sn(y) = y−xm
∑nk=1
1k
, and therefore
limn→∞ |Sn(x)− Sn(y)| = ∞,
which contradicts our assumptions.
Comments 82 If xm is the value for which S(x) converges, then S(xm) = logm.
Problem 4.19. Prove that if zi ∈ C, 0 < |zi | ≤ 1, then |z1 − z2| ≤ | log z1 − log z2|.Solution 4.19. From Lagrange’s mean value theorem, we have | log r1 − log r2| =1ξ|r1 + r2| for any 0 < r1, r2 < 1 and ξ ∈ [0, 1]. Thus | log r1 − log r2| ≤ |r1 − r2|.
Write zj = rj exp(iθj ). We have
(z1 − z2)2 = r2
1 + r22 − 2r1r2 cos(θ1 − θ2) = (r1 − r2)
2 + 2r1r2(1 − cos(θ1 − θ2))
= (r1 − r2)2 + 4r1r2 sin2 1
2(θ1 − θ2) ≤ (r1 − r2)
2 + r1r2(θ1 − θ2)2
≤ (r1 − r2)2 + (θ1 − θ2)
2 ≤ (log r1 − log r2)2 + (θ1 − θ2)
2
= | log z1 − log z2|2.
We have used above the inequality sin θ < |θ |, and its consequence 4 sin2 12 (θ1 −
θ2) ≤ (θ1 − θ2)2.
Problem 4.20. The roots of the function of a complex variable
ζ4(s) = 1 + 2s + 3s + 4s
are simple.
Solution 4.20. Suppose that ζ4(s) and ζ ′4(s) have common zeros. Let us set x = 2s .
Then we have
1 + x + x2 = −3s and x log 2 + x2 log 4 = −3s log 3,
so thatx2(log 3/2)+ x(log 3/2)+ log 3 = 0.
The discriminant of this quadratic equation is positive; hence the roots are real: 2s =x ∈ R, 3s = −(1 + x + x2) ∈ R.
It is known that s cannot be real, since ζ4 has no real zeros. However, the realitycondition above shows that the (nonzero) imaginary part of s must be an integermultiple of both πi
log 2 and πilog 3 . This is impossible, because log 3/ log 2 �∈ Q.
Comments 83 A related function that generalizes the finite sum ζ4(s) is the Riemannzeta function ζ defined by
ζ(s) =∑
k≥1
k−s .
8 Analysis Solutions 195
It is not hard to see that ζ(s) is defined for all complex s with real part �(s) > 1 andcan be continued analytically in s. The function ζ(s) has zeros at the negative evenintegers −2,−4, . . . and one usually refers to them as the trivial zeros.
One of the most important problems in mathematics today, known as the Riemannhypothesis, is to decide whether all nontrivial zeros of ζ(s) in the complex plane lie onthe “critical line” consisting of points of real part 1
2 . The nontrivial zeros correspondprecisely to the zeros of the following even entire function:
ξ(t) = π−s/2s(s − 1)
2�( s
2
)ζ(s),
where s = 12 + √−1t and � is the usual gamma function. This is known to hold true
for the first 1013 zeros and for 99 percent of the zeros. The validity of the conjectureis equivalent to saying that
π(n)−∫ n
0
dt
log t= O(√
n log n),
where π(n) counts the number of primes less than n.The Riemann hypothesis was included as one of the seven Millennium Problems
(for which the Clay Mathematics Institute will award one million dollar for a solution).The interested reader might consult:
• B. Conrey: The Riemann hypothesis, Notices Amer. Math. Soc. March, 2003,341-353.
• E. Bombieri: Problems of the millennium: The Riemann Hypothesis, The ClayMathematics Institute, 2000.http://www.claymath.org/millennium/Riemann_Hypothesis/
Problem 4.21. Compute
I =∫ π/2
0log sin x dx.
Solution 4.21. We have
I =∫ π/2
0log cos x dx =
∫ π
π/2log sin x dx = 1
2
∫ π
0log sin x dx
and thus
2I = 2∫ π/2
0log sin 2x dx
= 2
(∫ π/2
0log 2 dx +
∫ π/2
0log sin x dx +
∫ π/2
0log cos x dx
)= 4I + π log 2
and thus I = −π2 log 2.
196 8 Analysis Solutions
Problem 4.22. Letf a be positive, monotonically decreasing function on [0, 1]. Provethe following inequality:
∫ 10 xf
2(x)dx∫ 1
0 xf (x)dx≤
∫ 10 f
2(x)dx∫ 1
0 f (x)dx.
Solution 4.22. The above inequality is equivalent to
∫ 1
0f 2(x) dx
∫ 1
0yf (f ) dy −
∫ 1
0yf 2(y) dy
∫ 1
0f (x) dx ≥ 0,
which can be rewritten as follows:
I =∫ 1
0
∫ 1
0f (x)f (y)y(f (x)− f (y)) dx dy ≥ 0.
Interchanging the variables and the integration order, we obtain
2I =∫ 1
0
∫ 1
0f (x)f (y)(y − x)(f (x)− f (y)) dx dy ≥ 0.
Since f is decreasing, we have (y − x)(f (x) − f (y)) ≥ 0 and thus 2I ≥ 0, asclaimed.
Comments 84 It can be proven by the same method that
∫ 1
0f (x)g(x)w(x) dx
∫ 1
0w(x) dx ≤
∫ 1
0f (x)w(x) dx
∫ 1
0g(x)w(x) dx,
where f is decreasing, g is increasing, and w is nonnegative, with strict inequality iff and g are not constant on the support of w.
Problem 4.23. Prove that every real number x such that 0 < x ≤ 1 can be representedas an infinite sum
x =∞∑
k=1
1
nk,
where the nk are natural numbers such that nk+1nk
∈ {2, 3, 4}.
Solution 4.23. We have two cases to consider, 13 ≤ x ≤ 1 and 0 < x < 1
3 .In the first case, let us put n0 = 1 and x0 = x. We define the sequence xk =
nknk−1
xk−1 − 1, where
nk =
⎧⎪⎨
⎪⎩
4nk−1, if 13 ≤ xk−1 <
12 ,
3nk−1, if 12 ≤ xk−1 <
23 ,
2nk−1, if 23 ≤ xk−1 < 1.
8 Analysis Solutions 197
If 13 ≤ xk−1 ≤ 1, then it is immediate that 1
3 ≤ xk ≤ 1 and so, 13 ≤ xk ≤ 1. Further,
the sequence {nk} is unbounded, which implies that
limk→∞
xk
nk= 0.
The recurrence relation reads
xk
nk= xk−1
nk−1− 1
nk,
and hencexk
nk= x0
n0−
k∑
i=1
1
ni= x −
k∑
i=1
1
ni.
Thus by letting k go to infinity, x = ∑∞i=1
1ni
.
In the second case, 0 < x < 13 , we choose a natural number k ≥ 4 such that
1/k ≤ x < 1/(k − 1). Set N = 2(k − 1). Then the number x1 = Nx − 1 satisfies13 < x1 < 1. From the above,
x1 =∞∑
k=1
1
nk,
where nk+1nk
∈ {2, 3, 4}. Then we write
x = 1
N+ x1
N= 1
N+
∞∑
k=1
1
Nnk=
∞∑
k=1
1
mk,
where mk = Nnk−1, for k ≥ 2, and m1 = N .
Comments 85 In general, the representation is not unique, since for example,
1
2=
∞∑
k=2
1
3k=
∞∑
k=1
1
2k.
Moreover, the result of the problem holds as well for those x that belong to the interval[ 43 , 2
].
Problem 4.24. Let Sn denote the set of polynomials p(z) = zn + an−1zn−1 + · · · +
a1z+ 1, with ai ∈ C. Find
Mn = minp∈Sn
(max|z|=1
|p(z)|).
Solution 4.24. Let us show thatMn = 2 if n ≥ 2. We have firstMn ≤ max|z|=1 |zn+1| = 2.
Suppose now that Mn < 2, for some n, so that there exists a polynomial p(z) =zn + an−1z
n−1 + · · · + a1z + 1 with the property that max|z|=1 |p(z)| < 2. Thisimplies that
198 8 Analysis Solutions
−2 < �(p(z)) < 2, for z ∈ C, with |z| = 1.
Let ξj , j = 1, . . . , n be the nth roots of unity. Sum the inequalities above for z = ξj ,and get
−2n < �n∑
j=1
p(ξj ) < 2n.
On the other hand, an immediate computation shows that
n∑
j=1
p(ξj ) = 2n
for any p as above, which is a contradiction. Thus our claim follows.
Problem 4.25. Find an explicit formula for the value of Fn that is given by the recur-rence Fn = (n+ 2)Fn−1 − (n− 1)Fn−2 and initial conditions F1 = a, F2 = b.
Solution 4.25. We will use the so-called Laplace method. Letϑ : R → R be a smoothfunction such that
Fn =∫ β
α
tn−1ϑ(t) dt.
The recurrence satisfied by Fn is then a consequence of the following differentialequation:
dϑ
ϑ= 1 − 3t + t2
t − t2dt.
This differential equation can be integrated, and one obtains the solutionϑ = e−t t (t−1). Further, the integral’s limits α and β are among the roots of the equation
tn+1(t − 1)e−t = 0,
yielding t1 = 0, t2 = ∞, t3 = 1. The general solution is therefore
Fn = C1
∫ ∞
0tn(t − 1)e−t dt + C2
∫ 1
0tn(t − 1) e−tdt
= C1n!n+ C2
(
n!n
(
1 − 1
e
n∑
k=0
1
k!
)
− 1
e
)
= (3b − 11a)n!n+ (4a − b)
(
1 + n!nn∑
k=0
1
k!
)
.
Problem 4.26. Find the positive functions f (x, y) and g(x, y) satisfying the follow-ing inequalities:
( n∑
i=1
aibi)2 ≤ ( n∑
i=1
f (ai, bi))( n∑
i=1
g(ai, bi)) ≤ ( n∑
i=1
a2i
)( n∑
i=1
b2i
)
for all ai, bi ∈ R and n ∈ Z+.
8 Analysis Solutions 199
Solution 4.26.(ab)2 ≤ f (a, b)g(a, b) ≤ a2b2
and thus(1) f (a, b)g(a, b) = a2b2.
For n = 2, use the hypothesis
2a1b1a2b2 ≤ f (a1, b1)g(a2, b2)+ f (a2, b2)g(a1, b1) ≤ a21b
22 + a2
2b21
and the identity (1) to find that
(2) 2 ≤ f (a1, b1)
f (a2, b2)
a2b2
a1b1+ f (a2, b2)
f (a1, b1)
a1b1
a2b2≤ a1b2
a2b1+ a2b1
a1b2.
We replace (a1, b1) = (a, b) and (a2, b2) = (λa, λb); then
2 ≤ f (a, b)
f (λa, λb)λ2 + f (λa, λb)
f (a, b)λ−2 ≤ 2,
from which we derivef (λa, λb) = λ2f (a, b).
Set f (a) = f (a, 1), and so f (a, b) = b2f (a/b). Then (2) becomes
(3) 2 ≤ f (a)/a
f (b)/b+ f (b)/b
f (a)/a≤ a
b+ b
a.
This implies that when a ≥ b, we have f (a)bf (b)a
≤ ab
and f (b)af (a)b
≤ ab
. This yields
(4) f (b) ≤ f (a),f (a)
a2 ≤ f (b)
b2 , (a ≥ b).
Conversely, if f, g satisfy (1) and (4), then by applying the inequality (2) for allpairs of indices i �= j ≤ n, we find that
2∑
aibiaj bj ≤∑
i,j
[f (ai, bi)g(aj , bj )+f (aj , bj )g(ai, bi)] ≤∑
i,j
(a2i b
2j +a2
j b2i
),
and the inequality from the statement follows.
Comments 86 If f (x, y) = x2 + y2, g(x, y) = x2y2/(x2 + y2), then we obtain aninequality due to Hardy, Littlewood, and Pólya. By choosing f (x, y) = x1+αy1−α ,g(x, y) = x1−αy1+α , we obtain the Calderbank inequality.
Problem 4.27. Let g ∈ C1(R) be a smooth function such that g(0) = 0 and |g′(x)| ≤|g(x)|. Prove that g(x) = 0.
200 8 Analysis Solutions
Solution 4.27. Let x ≥ 0. We have
|g(x)| ≤∣∣∣∣
∫ x
0g′(t) dt
∣∣∣∣ ≤
∫ x
0|g′(t)| dt ≤
∫ x
0|g(t)| dt.
Let a > 0 be arbitrary. Since g ∈ C1(R), g is bounded on any bounded interval andthus there exists k, such that |g(x)| ≤ k, for 0 ≤ x ≤ a.
If 0 ≤ t ≤ x ≤ a, then |g(t)| ≤ k, and so
|g(x)| ≤∫ x
0k dt = kx
for every x with 0 ≤ x ≤ a. Now the condition |g(t)| ≤ kt , for 0 ≤ t ≤ x ≤ a,implies that
|g(x)| ≤∫ x
0kt dt = 1
2kx2, for 0 ≤ x ≤ a.
Continuing this way, we prove by induction that
|g(x)| ≤ 1
n!kxn, provided that 0 ≤ x ≤ a.
Putting x = a and passing to the limit n → ∞, we obtain
|g(a)| ≤ limn→∞
1
n!kan = 0.
Since a was arbitrary, we have g(x) = 0 for all x ∈ [0,∞). A similar argumentsettles the case x ∈ (−∞, 0].
Problem 4.28. Let a1, b1, c1 ∈ R+ such that a1 + b1 + c1 = 1 and define
an+1 = a2n + 2bncn, bn+1 = b2
n + 2ancn, cn+1 = c2n + 2anbn.
Prove that the sequences (an), (bn), (cn) have the same limit and find that limit.
Solution 4.28. Let
un = an + bn + cn, vn = an + ωbn + ω2cn, wn = an + ω2bn + ωcn,
where ω3 = 1 is a third root of unity. The given recurrence for an, bn, cn induces thefollowing recurrence relations:
un+1 = u2n = 1, vn+1 = w2
n, wn+1 = v2n.
Thus wn+1 equals either v2n1 or w2n
1 , depending on the parity of n (and similarlyfor vn+1). Since v1 and w1 are convex combinations with positive coefficients ofthe complex points 1, ω, ω2, it follows that |v1|, |w1| < 1. But then we derive thatlimn→∞ vn = limn→∞wn = 0. Since un is constant, we obtain
8 Analysis Solutions 201
limn→∞ an = lim
n→∞1
3(un + vn + wn) = 1
3,
limn→∞ bn = lim
n→∞1
3(un + ω2vn + ωwn) = 1
3,
limn→∞ cn = lim
n→∞1
3(un + ωvn + ω2wn) = 1
3.
Problem 4.29. Consider the sequence (an), given by a1 = 0, a2n+1 = a2n = n− an.Prove that an = n
3 , for infinitely many values of n. Does there exist an n such that|an − n
3 | > 2005? Also, prove that
limn→
an
n= 1
3.
Solution 4.29. 1. We have an = [n2
] − a[ n2
], and thus
an =[n
2
]−
[n4
]+
[n8
]−
[ n10
]+ · · · .
If n = 3 · 2k , then we have
a3·2k = 3 · 2k
2− 3 · 2k
4+ · · · + 3 · 2k
2k−
[3
2
]
= 3(2k−1 + (−2k−2)+ 2k−3 + (−2k−4) · · · )− 1
= 3 · 2k + 1
3− 1 = 2k,
and therefore an = n3 .
2. Set nk = 4k−13 . Then nk is odd and satisfies the recurrence nk − 1 = 4nk−1.
This implies that
ank = ank−1 = a4nk−1 = 2nk−1 − a2nk−1 = 2nk−1 − nk−1 + ank−1 = nk−1 + ank−1
and thus
ank = nk−1 + nk−2 + · · · + n1 + a1 = 1
3
(4k − 1
3− k
).
Thus∣∣ank − nk
3
∣∣ = k3 , and so
∣∣ank − nk3
∣∣ can be arbitrarily large.
3. Use the fact that n2k
− 1 ≤[n2k
]≤ n
2kand find that
⎛
⎝[log n]∑
k=1
(−1)kn
2k
⎞
⎠ − [log n] ≤ an ≤⎛
⎝[log n]∑
k=1
(−1)kn
2k
⎞
⎠ + [log n] + 1,
and after dividing by n and letting n go to infinity, we obtain the claim.
202 8 Analysis Solutions
Problem 4.30. Compute the integral
f (a) =∫ 1
0
log(x2 − 2x cos a + 1)
xdx.
Solution 4.30.
f(a
2
)+ f
(π − a
2
)=
∫ 1
0
log(x2 − 2x cos( a2 )+ 1)(x2 + 2x cos(π − a2 )+ 1)
xdx
=∫ 1
0
log(x4 − 2x2 cos a + 1)
xdx.
Make the variable change x = √t and find that the last integral equals
1
2
∫ 1
0
(log t2 − 2t cos a + 1)
tdt = f (a)
2, for a ∈ [0, 2π ].
Therefore, f satisfies the identity
f(a
2
)+ f
(π − a
2
)= f (a)
2.
By differentiating twice (we have f is C2), we obtain
f ′′ (a2
)+ f ′′ (π − a
2
)= 2f ′′(a).
Suppose that f ′′(a) has the maximumM and the minimumm, and let a0 be such that
f ′′(a0) = M, a0 ∈ [0, 2π ].
Putting a = a0 in the identity above we get
f ′′ (a0
2
)+ f ′′ (π − a0
2
)= 2f ′′(a0) = 2M,
from which we obtain f ′′(a0/2) = M . Iterating this procedure, we find that f ′′(a0) =f ′′ ( a0
2n) = M , for any M . From the continuity of f ′′, we find that
limn→0
f ′′ (a0
2n
)= f ′′(0) = M.
But now a similar argument dealing with the minimum shows that f ′′(0) = m andtherefore M = m; hence f ′′ is constant. This implies that f (a) = αa2/2 + βa + γ .Substituting in the identity above, we obtain the following identities in the coefficients:
−πα = β, π2α/2 + βπ + 2γ = γ /2.
Using now f ′(π2 ) = π2 , it follows that πα/2 + β = π/2, therefore α = −1, β =
π, γ = −π2/3. Thus we obtain
f (a) = −a − 2
2+ πa − π2
3.
8 Analysis Solutions 203
Problem 4.31. Let −1 < a0 < 1, and define an = ( 12 (1 + an−1)
)1/2for n ≥ 1. Find
the limits A,B, and C of the sequences
An = 4n(1 − an), Bn = a1 · · · an, Cn = 4n(B − a1a2 · · · an).Solution 4.31. Let a0 = cos θ . By making use of the formula cos2 θ/2 = (1 +cos θ)/2 and a recurrence on n, we derive that an = cos(θ/2n). Then
An = 4n(1−an) = 4n(θ2
2 · 4n− θ4
24 · 42n + O(4−3n)
)= θ2
2− θ4
24 · 4n+O(
4−2n).
Thus A = θ2
2 . Further, sin(θ/2n)Bn = cos(θ/2) cos(θ/22) · · · cos(θ/2n) sin(θ/2n)and thus
Bn = sin θ
2n sin(θ/2n)= sin θ
θ+ sin θ
6 · 4n+ O(4−2n)
and thus B = sin θθ
. Finally, Cn = −θ sin θ6 + O(
4−n), and so C = −θ sin θ6 .
Comments 87 The same method works when a0 > 1, by making use of the hyperbolictrigonometric functions. If we put
θ = log(a0 +
√a2
0 − 1) = arccosh a0,
then A = −θ2/2, B = sinh θ6 , C = −6 sinh θ
6 .
Problem 4.32. Consider the sequence given by the recurrence
a1 = a, an = a2n−1 − 2.
Determine those a ∈ R for which (an) is convergent.
Solution 4.32. For |a| > 2, we have |an| > 2 and consequently an+1 > an for alln ≥ 2. Since the sequence is increasing, it is either convergent or tends to ∞. Ifwe have a finite limit an → λ, then it satisfies λ = λ2 − 2 and thus λ ∈ {−1, 2},contradicting our assumptions.
Therefore, we need |a| ≤ 2 for the convergence. Let α ∈ (−π2 ,
π2
)be such that
2 cosα = a. By induction on n, we obtain
an = 2 cos 2nα.
Assume that α is not commensurable with π , that is, απ/∈ Q. Observe that the
function ϕ(x) = 2xα is continuous. Thus, the map ϕ : R/πZ → R/πZ sends thedense subset {nα (modπ), n ∈ Z} into a dense subset, which is {2nα (modπ), n ∈ Z}.Therefore, the image under cos, namely {cos 2nα; n ∈ Z}, is dense in [−1, 1], andthus the sequence (an) cannot be convergent.
If α/π = p/q, where p, q ∈ Z, then the sequence (an) is periodic for n largeenough. Thus, in order for the sequence to be convergent it is necessary and sufficientthat it be constant for n large enough. This condition reads
204 8 Analysis Solutions
cos(2nα
) = cos(2n+1α
).
Thus2nα = ±2n+1α + kπ, where k ∈ Z,
and soα ∈
{ m
2n3π, where m, n ∈ Z
}.
One sees that an is convergent and constant for n ≥ m.
Problem 4.33. Let 0 < a < 1 and I = (0, a). Find all functions f : I → R
satisfying at least one of the conditions below:
1. f is continuous and f (xy) = xf (y)+ yf (x).2. f (xy) = xf (x)+ yf (y).
Solution 4.33. 1. Let g(x) = f (x)x
. Then g(exp(x)) is continuous and additive onI . This implies that g(exp(x)) is linear and hence g(x) = C log x, which impliesf (x) = Cx log x for some constant C.
2. Give y the values x, x2, x3, and x4. We find therefore that
f(x2) = 2xf (x),
f(x3) = xf (x)+ x2f
(x2) = (
x + 2x3)f (x),
f(x4) = xf (x)+ x3f
(x3) = (
x + x4 + 2x6)f (x),
f(x4) = x2f (x)+ x2f
(x2) = 4x3f (x).
From the last two lines, one derives that f (x) = 0 for all but those points x for whichx + x4 + 2x6 = 4x3. This means that f vanishes for all but finitely many (actually6) values of x.
If t ∈ I is such that f (t) �= 0, then the first identity shows that f(t2) �= 0.
By induction on n we obtain f (t2n) �= 0. This means that f does not vanish for
infinitely many values of the argument, which contradicts our former result. Thus fis identically zero.
Problem 4.34. If a, b, c, d ∈ C, ac �= 0, prove that
max(|ac|, |ad + bc|, |bd|)max(|a|, |b|)max(|c|, |d|) ≥ −1 + √
5
2.
Solution 4.34. Set ab−1 = r, dc−1 = s, −1+√5
2 = k, k2 = 1 − k, 0 < k < 1. Theinequality from the statement is equivalent to
μ = max(1, |r + s|, |rs|) ≥ kmax(1, |r|)max(1, |s|) = ν.
(i) If |r| ≥ 1, |s| ≥ 1, then
μ ≥ |rs| > k|s||r| = ν,
8 Analysis Solutions 205
proving our claim.(ii) If |r| ≤ 1, |s| ≥ 1, then our inequality is equivalent to
max(1, |r + s|, |rs|) ≥ k|s|.Moreover, if k|s| ≤ 1 or |r + s| ≥ k|s|, then this is obviously true.
Further, let us assume that k|s| > 1 and |r+s| < k|s|. We have then |r|+|r+s| ≥|s| and consequently,
|r| ≥ |s| − |s + r| ≥ |s| − k|s| = (1 − k)|s| = k2|s|.Furthermore, |rs| ≥ k2|s|2 > k|s| > 1; thus max(1, |r + s|, |rs|) ≥ k|s| holds.
(iii) If |r| ≥ 1, |s| ≤ 1, then the inequality is proved as in case (ii), by using theobvious symmetry.
Problem 4.35. Let∑∞i=1 xi be a convergent series with decreasing terms x1 ≥ x2 ≥
· · · ≥ xn ≥ · · · > 0 and let P be the set of numbers which can be written in the form∑i∈J xi for some subset J ∈ Z+. Prove that P is an interval if and only if
xn ≤∞∑
i=n+1
xi for every n ∈ Z+.
Solution 4.35. 1. Suppose that there exists p such that xp >∑∞i=p+1 xi and consider
α such that∑i>p xi < α < xp. Set S(J ) = ∑
i∈J xi . We claim that there doesnot exist any J ⊂ Z+ such that S(J ) = α. In fact, if J ∩ {1, 2, . . . , p} �= ∅, then∑i∈J xi ≥ xp > α, becausexk is decreasing. On the other hand, ifJ∩{1, 2, . . . , p} =
∅, then∑i∈J xi ≤ ∑
i>p xi < α. Finally, P contains numbers smaller than α, forinstance
∑i>p xi , as well as numbers bigger than α, such as xp. Thus P is not an
interval.2. Assume now that the hypothesis of the problem is satisfied and choose any
element y from the interval (0, S], where S = ∑∞i=1 xi . We will show that there
exists L such that S(L) = y.Let n1 be the smallest index such that xn1 < y. There exists such a one because
limk→∞ xk = 0. By induction, we define nk to be the smallest integer with theproperty that
xn1 + · · · + xnk < y.
Let L = {n1, . . . , nk, . . . }.It is clear that S(L) ≤ y. If p ∈ Z+ \ L, then choose the smallest k with the
property that nk > p. Since p does not belong to L, we have
xn1 + · · · + xnk−1 + xp ≥ y.
This implies that ( ∞∑
i∈Lxi
)
+ xp ≥ y.
206 8 Analysis Solutions
(i) Assume that the set Z+ − L is finite. This set is empty only when y = S,which obviously belongs to P . Now take p to be the maximal element of Z+ − L.This implies that all elements bigger than p already belong to L and hence
L = {n1, . . . , nk−1} ∪ {p + 1, p + 2, . . . }.Then
∑
i∈Lxi =
k−1∑
i=1
xn1 +∞∑
j=p+1
xj ≥k−1∑
i=1
xni + xp ≥ y,
from which we obtain that∑i∈L xi = y.
(ii) If Z+ − L is infinite, then for any ε > 0, one can choose some pε ∈ Z+ − L
such that xpε < ε. The inequality above for pε implies that
ε +∑
i∈Lxi ≥ y.
This is true for any positive ε, and thus∑i∈L xi = y.
Problem 4.36. Does there exist a continuous function f : (0,∞) → R such thatf (x) = 0 if and only if f (2x) �= 0? What if we require only that f be continuous atinfinitely many points?
Solution 4.36. 1. There exists some real number α such that f (α) = 0, and thusf (2α) �= 0. Let γ ∈ [α, 2α] be given by γ = sup{x ∈ [α, 2α] such that f (x) = 0}.Since f is continuous, we find that f (γ ) = 0, and thus γ < 2α.
Moreover, if xn ∈ R is a sequence such that limn xn = γ , then f (xn) = 0 for nlarge enough. In fact, otherwise there exists a subsequence f (xnk ) �= 0, which impliesthat f (2xnk ) = 0, and thus limk f (2xnk ) = 0 �= f (2γ ).
Finally, recall that γ is the supremum of those x ∈ [α, 2α] for which f (x) = 0 andγ < 2α. Thus for n large such that γ + 1
n< 2α, there exists xn with γ < xn < γ + 1
nfor which f (xn) �= 0. This contradicts the previous claim. Thus there does not exista continuous f as in the statement.
2. The function
f (x) ={
0, for 22k < x ≤ 22k+1,where k ∈ Z,
1, for 22k−1 < x < 22k,where k ∈ Z,
satisfies the claim and has a countable number of discontinuities.
Problem 4.37. Find the smallest number a such that for every real polynomial f (x)of degree two with the property that |f (x)| ≤ 1 for all x ∈ [0, 1], we have |f ′(1)| ≤ a.Find the analogous number b such that |f ′(0)| ≤ b.
Solution 4.37. Let f (x) = ux2 + vx + w. We have
|f ′(1)| = |2u+ v| = |3f (1)− 4f (1/2)+ f (0)| ≤ 3|f (1)| + 4|f (1/2)| + |f (0)|≤ 3 + 4 + 1 = 8.
8 Analysis Solutions 207
Further, a = 8 because we have equality above for f (x) = 8x2 − 8x + 1.For the second case, we observe that
|f ′(0)| = |v| = |4f (1/2)− 3f (0)− f (1)| ≤ 8
and thus b = 8 by taking f (x) = 8x2 − 8x + 1.
Problem 4.38. Let f : R → R be a function for which there exists some constantM > 0 satisfying
|f (x + y)− f (x)− f (y)| ≤ M, for all x, y ∈ R.
Prove that there exists a unique additive function g : R → R such that
|f (x)− g(x)| ≤ M, for all x ∈ R.
Moreover, if f is continuous, then g is linear.
Solution 4.38. Let us show first the existence of the upper limit ϕ(x) =lim supn→∞
f (nx)n
.In fact, using the hypothesis, one finds that f (nx) ≤ n(f (x) + M), and hence
the sequence f (nx)n
is bounded from above, and thus ϕ(x) exists.Furthermore,
∣∣∣∣f (n(x + y))
n− f (nx)
n− f (ny)
n
∣∣∣∣ ≤ M
n,
which yields ϕ(x)+ ϕ(y) = ϕ(x + y); thus ϕ is additive.Next, observe that f (nx) ≤ n(f (x) +M) implies that
∣∣nxn
− f (x)∣∣ ≤ M , and
we can take g(x) = ϕ(x).If f is continuous, then it is easy to see that the lower limit, lim infn→∞ f (nx)
n,
coincides with ϕ(x), and moreover, ϕ(x) is continuous. An additive continuous func-tion is therefore linear. In fact, we have f (nx) = nf (x) for n ∈ Z, which yieldsf (rx) = rf (x) for any r ∈ Q. Since f is continuous and every real number r canbe approximated by a sequence of rational numbers, we find that f (rx) = rf (x) forany r ∈ R, and hence the claim.
Let us prove the uniqueness of such additive functions. Assume that there existsanother additive function h satisfying
|f (x)− h(x)| ≤ M, for all x ∈ R.
This implies that|g(x)− h(x)| ≤ M, for all x ∈ R.
However, an additive function is either zero or unbounded. Therefore, the additivefunction g − h must vanish.
208 8 Analysis Solutions
Comments 88 Functions f satisfying the condition from the statement are calledquasimorphisms.
Notice that there exist additive functions ϕ : R → R that are not linear (not contin-uous of course). Examples can be constructed by choosing a basis of R as a vectorspace over Q and defining arbitrarily the values of the function on each vector of thatbasis.
Problem 4.39. Show that if f is differentiable and if
limt→∞
(f (t)+ f ′(t)
) = 1,
thenlimt→∞ f (t) = 1.
Solution 4.39. Let us show first that if limt→∞(f ′(t) + αf (t)) = 0, where a =�(α) > 0, then limt→∞ f (t) = 0. Let ε > 0 and c < ∞ be such that |f ′(t) +αf (t)| ≤ aε whenever t ≥ c. Then
∣∣∣∣d
dt
(eαtf (t)
)∣∣∣∣ = ∣∣eαt (f ′(t)+ αf (t))
∣∣ ≤ aεeat
for t ≥ c. Therefore, using the mean value theorem we have
|eαtf (t)− eacf (c)| ≤ ε(eαt − eac
), for t > c.
This yields|f (t)| ≤ ea(c−t) · |f (c)| + ε
∣∣1 − ea(c−t)∣∣.
In particular, for sufficiently large t , we have |f (t)| ≤ 2ε, as claimed. Now thestatement follows by applying this result to f − 1.
Comments 89 Let P(D) be a polynomial in the derivation operator D = ddt . Let us
writeP(D) = (D − λ1)(D − λ2) · · · (D − λn).
The above result can be generalized as follows. Assume that �λi < 0, where � denotesthe real part; if the function f satisfies limt→∞ P(D)f (t) = 0, then limt→∞ f (t) =0. This is proved by recurrence on the degree of P . The recurrence step follows fromthe argument given above. In particular, the claim holds for P(D) = D2 + D + 1.
The condition �λi < 0 is necessary. In fact, we can consider f (t) = eλi t , whichsatisfies limt→∞ P(D)f → 0. Also, one notices that P(D) = D0 + D1 + ... + Dn
does not fulfill this condition for n ≥ 3.
Problem 4.40. Let c be a real number and letf : R → R be a smooth function of classC3 such that limx→∞ f (x) = c and limx→∞ f ′′′(x) = 0. Show that limn→∞ f ′(x) =limx→∞ f ′′(x) = 0.
8 Analysis Solutions 209
Solution 4.40. According to Taylor’s formula, there exist ξx, ηx ∈ (0, 1) such that
f (x + 1) = f (x)+ f ′(x)+ 1
2f ′′(x)+ 1
6f ′′′(x + ξx),
f (x − 1) = f (x)− f ′(x)+ 1
2f ′′(x)− 1
6f ′′(x + ηx).
Suitable linear combinations of these yield
f ′′(x) = f (x + 1)− 2f (x)+ f (x − 1)− 1
6f ′′′(x + ξx)+ 1
6f ′′′(x − ηx),
2f ′(x) = f (x + 1)− f (x − 1)− 1
6f ′′′(x + ξx)− 1
6f ′′′(x − ηx).
Since x+ ξx → ∞ and x−ηx → ∞ as x → ∞, by passing to the limit, one obtains
limx→∞ f
′′(x) = c − 2c + c − 1
6· 0 + 1
6· 0 = 0,
limx→∞ f
′(x) = 1
2
(c − c − 1
6· 0 − 1
6· 0
)= 0.
Comments 90 It can be proved that if limx→∞ f (x) exists and if f (n) is bounded,then limx→∞ f (k)(x) = 0 for any 1 ≤ k < n. See for instance:
• J. Littlewood: The converse of Abel’s theorem, Proc. London Math. Soc. (2) 9(1910–11), 434–448.
Problem 4.41. Prove that the following integral equation has at most one continuoussolution on [0, 1] × [0, 1]:
f (x, y) = 1 +∫ x
0
∫ y
0f (u, v) du dv.
Solution 4.41. If f1, f2 are two solutions, then their difference g satisfies
g(x, y) =∫ x
0
∫ y
0g(u, v) du dv.
Since g is continuous, its absolute value is bounded by some constantM on the square[0, 1] × [0, 1]. Thus
|g(x, y)| ≤∫ x
0
∫ y
0M du dv = Mxy.
By induction, we show that |g(x, y)| ≤ Mxnyn
(n!)2. The induction step follows from
|g(x, y)| ≤∫ x
0
∫ y
0
Munvn
(n!)2du dv = M(xy)n+1
((n+ 1)!)2.
If we fix x, y, then
0 ≤ |g(x, y)| ≤ limn→
Mxnyn
n!n!= 0,
and thus g(x, y) ≡ 0.
210 8 Analysis Solutions
Comments 91 There exists a continuous solution of this equation, given by
f (x, y) = 1 + xy + x2y2
2!2!+ x3y3
3!3!+ · · · = J0
(√−4xy) = J0
(2i
√xy
),
where J0 is the Bessel function of order zero.
Problem 4.42. Find those λ ∈ R for which the functional equation
∫ 1
0min(x, y)f (y) dy = λf (x)
has a solution f that is nonzero and continuous on the interval [0, 1]. Find thesesolutions.
Solution 4.42. The equation can be written in the form
λf (x) =∫ x
0yf (y) dy + x
∫ 1
x
f (y) dy.
If λ �= 0, then f is differentiable, and after derivation, we obtain
λf ′(x) = xf (x)− xf (x)+∫ 1
x
f (y)dy =∫ 1
x
f (y)dy.
Thus f ′ is differentiable and
λf ′′(x) = −f (x).This implies that f (x) = A cosμx + B sinμx, where μ = λ−1/2 if λ > 0, andf (x) = A cosh νx + B sinh νx, where ν = (−λ)−1/2 if λ < 0. According to thehypothesis,
limn→0
f (x) = 0,
and hence A = 0 for any choice of λ. From the above, we also get
limn→0
f ′(x) = 0,
and thus Bμ cosμ = 0 and Bν cosh ν = 0. The last equation yields B = 0, and sowe do not have nonzero solutions. For λ > 0, if B �= 0, then cosμ = 0 and thusμ = (2k + 1)π2 . Thus λ = μ−2 4
(2k+1)2π2 , where k ∈ Z, and
f (x) = B cos(2k + 1)π
2x, B ∈ R.
If λ = 0, the same method gives us f (x) = 0.
8 Analysis Solutions 211
Comments 92 We proved that the integral operator T : C0(0, 1) → C0(0, 1) definedon the space C0(0, 1) of continuous functions on (0, 1) and given by the formula
(Tf )(x) =∫ 1
0min(x, y)f (y)dy
has the eigenvalues 4(2k+1)2π2 and the eigenvectors cos(2k + 1)π2 x.
Problem 4.43. Let X be an unbounded subset of the real numbers R. Prove that theset
AX = {t ∈ R; tX is dense modulo 1}is dense in R.
Solution 4.43. Let {On}n∈Z+ be a countable base of (0, 1), On �= ∅. Let p be thecanonical projection p : R → R/Z. Define
An = {r ∈ R;p(rX) ∩On �= ∅}.This amounts to
An =⋃
x∈X−{0}
1
xp−1({On}),
and thus An is open. If F is an interval of length f > 0, then there exists x ∈ X
such that xf > 1, which implies that p(xF) = R/Z and thus there exists r ∈ F withp(rx) ∈ On. This proves that An is dense in R. Since
AX =∞⋂
n=0
An,
we obtain that AX is also dense in R, as the intersection of open dense subsets.
Problem 4.44. ConsiderP(z) = zn+a1zn−1 +· · ·+an, where ai ∈ C. If |P(z)| = 1
for all z satisfying |z| = 1, then a1 = · · · = an = 0.
Solution 4.44. Consider the polynomialQ(z) = 1 + a1z+ · · · + anzn, which can bewritten as Q(z) = P(z), when |z| = 1. This implies that |Q(z)| = 1 for all z of unitmodulus. Moreover,Q(0) = 1, and soQ has a value in the interior of the disk whosemodulus is equal to its maximum on the boundary circle. According to the maximumprinciple for holomorphic functions, Q has to be constant, Q(z) = 1, whence theclaim.
Comments 93 W. Blaschke has studied the complex analytic (i.e., holomorphic) func-tions f : D → C on the unit disk D that are continuous on D and have |f (z)| = 1for all z satisfying |z| = 1. He has shown that such a function has the form
f = σ
n∏
k=1
z− bk
1 − bkz,
where σ is a constant of modulus one and bk ∈ C are such that |bk| < 1.
212 8 Analysis Solutions
Problem 4.45. Let I ⊂ R be an interval and u, v : I → R smooth functions satisfy-ing the equations
u′′(x)+ A(x)u(x) = 0, v′′(x)+ B(x)v(x) = 0,
where A,B are continuous on I and A(x) ≥ B(x) for all x ∈ I . Assume that v is notidentically zero. If α < β are roots of v, then there exists a root of u that lies withinthe interval (α, β), unless A(x) = B(x), in which case u and v are proportional forα ≤ x ≤ β.
Solution 4.45. The roots of v are isolated and we can suppose that v|(α, β) > 0.Assume that u|(α, β), so that after a possible sign change, u|(α, β) > 0. Also, wehave v′(α) ≥ 0, v(β) ≤ 0, and since v is not identically zero, then v and v′ do nothave common roots; thus v′(α) > 0 and v′(β) < 0. We consider
w(x) = u(x)v′(x)− u′(x)v(x).
We have
w′(x) = u(x)v′′(x)− u′′(x)v(x) = (A(x)− B(x))uv ≥ 0,
and therefore w is increasing; so w(α) ≤ w(β) and thus u(α)v′(α) ≤ u(β)v′(β).Since u(α) ≥ 0, u(β) ≥ 0, v′(α) > 0, v′(β) < 0, we obtain u(α) = u(β) = 0, andsow(α) = w(β) = 0. Noww is increasing and hencew ≡ 0 and thusw′ ≡ 0, whichyields A(x) = B(x). Moreover, for α < x < β we have w(x)
v2(x)= ( u
v)′, and hence u
v
is constant.
Comments 94 This result is known as Sturm’s comparison theorem in differentialequations.
Problem 4.46. Let V be a finite-dimensional real vector space and f : V → R acontinuous mapping. For any basis B = {b1, b2, . . . , bn} of V , consider the set
EB = {z1b1 + · · · + znbn, where zi ∈ Z}.Show that if f is bounded on EB for any choice of the basis B, then f is boundedon V .
Solution 4.46. We solve first the case n = 1. Given any pair of positive numbers0 < a < b, there exists some r = r(a, b) (depending on a, b) such that for anyx > r , one can find m ∈ Z, with ma ≤ x ≤ mb. In fact, the intervals (ma,mb), forintegral m, will cover all of R but a compact set. Thus, for any x > r , one can findan interval (c, d) containing x, and some m ∈ Z such that a < c
m< d
m< b.
Assume that f is unbounded. Let us consider an interval I1 = [a1, b1] with theproperty that f (x) > 1, for x ∈ I1. Since f is unbounded, there exist x1 > r(a1, b1)
arbitrarily large such that f (x1) > 4 and m1 ∈ Z such that m1a1 < x1 < m1b1.Moreover, f is continuous and so f (x) > 2 for any x ∈ [c1, d1], for some smallinterval around x1, which can be chosen to be contained in (m1a1,m1b1). Set I2 =
8 Analysis Solutions 213
{x ∈ I1;m1x ∈ [c1, d1]}. Thus I2 ⊂ I1 is a compact nonempty interval with theproperty that f (m1x) > 2 for any x ∈ I2.
We define in this way a nested sequence of compact intervals I1 ⊂ I2 ⊂ I3 ⊂· · · Ik ⊂ · · · , and a sequence of integers m1 < m2 < m3 < · · · with the propertythat f (x) > 2k for x ∈ mkIk . This nested sequence of nontrivial compact intervalsmust have nontrivial intersection. Let a ∈ ∩∞
k=1Ik . Then the restriction of f to aZ isunbounded, since for any k, there exists some mk with f (mka) > 2k , by making useof the fact that a ∈ Ik .
Let now deal with the general case. Let zk ∈ V be a sequence for which f (zk) isunbounded, e.g., f (zk) > 2k+1. Then zk is unbounded, because f is continuous.
Let B = {b1, . . . , bn} be a basis of B and write zk = ∑ni=1 cikbi in the basis B.
There exists at least one i0 ∈ {1, 2, . . . , n} such that the sequence ci0k is unbounded;otherwise, zk would be bounded. One can slightly modify the basis B so that thiscondition is satisfied for all i. In fact, assume that cik is unbounded for 1 ≤ i ≤ m < n
and cik is bounded for m+ 1 ≤ i ≤ n. Take then the new basis vectors
b′1 = b1 −
n∑
i=m+1
bi, b′i = bi, when i ≥ 2.
In the new basis, one can express the vector zk as
zk =m∑
i=1
cikb′i +
n∑
i=m+1
(cik + c1k)b′i ,
and all components are now unbounded.We consider now the intervals Jik with the property that Uk = {x; x =∑ni=1 αikbi, with αik ∈ Jik} is a neighborhood of zk and f (x) > 2k , for all x ∈ Uk .The argument for n = 1 shows that there exists a sequence of integers (mik) going
to infinity such that we have a nested sequence of nontrivial compact intervals
· · · ⊂ 1
mik+1Jik+1 ⊂ 1
mikJik ⊂ · · · ⊂ Ji1.
Take
γi ∈∞⋂
k=1
1
mikJik
and construct a new basis B = {γibi, i ∈ {1, 2, . . . , n}}. It follows thatf is unboundedon EB .
Problem 4.47. It is known that if f, g : C → C are entire functions without commonzeros then there exist entire functionsa, b : C → C such thata(z)f (z)+b(z)g(z) = 1for all z ∈ C.
1. Prove that we can choose a(z) without any zeros.2. Is it possible to choose both a and b without zeros?
214 8 Analysis Solutions
Solution 4.47. 1. We know that Af + Bg = 1 for some entire functions A and B.Therefore, for any entire function λ, we have also
(A+ λg)f + (B − λf )g = 1.
We now choose a = A+λg such thatA+λg = eh. If z0 is a zero of g, we need to haveeh(z0) = A(z0) (with the same multiplicity). This is possible from the interpolationtheorem for entire functions, because A(z0)f (z0) = 1.
2. Let f, g ∈ C[z] be nonconstant polynomials of different degrees. We will showthat there are no entire functions a, b such that af + bg = 1 and a, b have no zeros.There are no integer nonconstant functions.
First, there exist nonzero polynomials A and B such that Af + Bg = 1. Asabove, for any entire function λ : C → C, one can construct other solutions, namelya = A − λg and b = B + λf . Let us now show that any pair a, b can be obtainedin this way. We have (A − a)f + (B − b)g = 0. Thus the set of zeros of (A − a)
contains the set of zeros of g (since f and g have no common zeros). In particular,there exists an entire function λ such that A− a = λg. In a similar way, there existsan entire function μ such that B − b = μf . The condition above shows that λ = μ,as claimed.
If a, b do not have zeros, then the meromorphic function F = 1λ
must be distincteverywhere from the rational functions a1, a2, a3 given below:
a1(z) = 0, a2(z) = g(z)/A(z), a3(z) = −f (z)/B(z).Consider the meromorphic function
G = F − a1
F − a2· a3 − a2
a3 − a1.
The points whereG(z) ∈ {0, 1,∞} are among those satisfying one of the conditionsa1(z) = a2(z), a1(z) = a3(z), a2(z) = a3(z), and therefore they are finitely many.According to Picard’s theorem,G does not have an essential singularity at ∞ and thusG is rational. This implies that F is rational; hence λf is a polynomial and thus a andb are polynomials. But polynomials without zeros are constant, and this is impossiblesince f and g have different degrees.
Problem 4.48. Consider a compact set X ⊂ R. Show that a necessary and sufficientcondition for the existence of a monic nonconstant polynomial with real coefficientsh ∈ R[x] such that |h(x)| < 1 for all x ∈ X is the existence of monic nonconstantpolynomial g(x) ∈ R[x] such that |g(x)| < 2 for all x ∈ X. Prove that 2 is themaximal number with this property.
Solution 4.48. Let P(X) be the Banach space of real polynomials endowed with thenorm ‖θ‖ = supx∈X |θ(x)|. We define the nonlinear operator
A : P(x) → P(x)given by
8 Analysis Solutions 215
(Af (x)) = f 2(x)− 1
2‖f ‖2, for f ∈ P(X).
If f is monic, thenAf is also monic and we have ‖Af ‖ = 12‖f ‖2. An easy induction
implies that
‖Anf ‖ = 2‖f ‖2n
22n .
Moreover, if ‖f ‖ < 2, then there exists some n for which ‖Anf ‖ < 1. This provesthe first part.
Let X = [−2, 2]. The norm of p(x) = x on X is ‖p‖ = 2. We will prove thatany monic polynomial on X has norm at least 2. Assume the contrary, namely thatthere exists f ∈ P(x), monic, with ‖f ‖ < 2. We have two intermediate results:
1. There exists an operator T : P(X) → P(X) that takes monic polynomials ofdegree 2k to monic polynomials of degree k such that
‖T P ‖ < ‖P ‖.2. Assuming the existence of a monic polynomial of norm ‖f ‖ < 2, there exists
n ∈ Z+ and a monic polynomial g, of degree 2n, with ‖g‖ < 2.By points 1 and 2 above there exists a monic polynomial g of degree 2n with
‖g‖ < 2 on [−2, 2], which implies that there exists a monic polynomial of degree 1such that ‖g‖ < 2 on [−2, 2], which is obviously false.
Proof of claim 1. If h ∈ P(x), let us consider h(x) = 12 (h(x) + h(−x)), which
is even and thus can be written as a polynomial in x2. Thus h(x) = Q(x2
), where
now Q : [0, 4] → R. Define T (h)(x) = Q(x + 2). It is clear that T has the desiredproperties.
Proof of claim 2. If f has degree 2mp, where p is odd, then t = T mf is monic ofdegreep, ‖t‖ < 2. There existsn ∈ Z+ such thatp divides 2n−1, and so 2n−1 = pq.We define g ∈ P(x) by means of g = t (x)qx. It is immediate that ‖g‖ < 2, and theproof is complete.
9
Glossary
9.1 Compendium of Triangle Basic Terminology and Formulas
9.1.1 Lengths in a Triangle
We used the standard notation for the important features of a triangle, as follows:
• a, b, c the sides lengths• A,B,C the angles• ma,mb,mc the medians• ha, hb, hc the altitudes• wa,wb,wc the angle bisectors• R the radius of the circumcircle• r the radius of the incircle• ra, rb, rc the radii of the extrinsic circles• S the area• p the semiperimeter
We collect below some useful identities between these quantities:
• S = 12bc sinA = (p(p − a)(p − b)(p − c))1/2 = 4Rr cos A2 cos B2 cos C2 ,
• S = rp,• 4RS = abc,• ra(p − a) = rp = S, ra = p tan A
2 ,• 4R + r = ra + rb + rc,• aha = 2S,• hahbhc = 8S3
abc= 2S2
R,
• 1ha
+ 1hb
+ 1hc
= 1ra
+ 1rb
+ 1rc
= 1r,
• m2a +m2
b +m2c = 3
4
(a2 + b2 + c2
),
• wa = 2√bc
b+c√p(p − a),
• ra =√p(p−b)(p−c)
p−a ,
218 9 Glossary
• a2 + b2 + c2 = 8R2(1 + cosA cosB cosC),
• sin A2 =
√(p−b)(p−c)
bc,
• cosA+ cosB + cosC = 1 + 4 sin A2 sin B
2 sin C2 = 1 + r
R,
• tan A2 =
√(p−b)(p−c)p(p−a) ,
• R = a2 sinA = b
2 sinB = c2 sinC ,
• 4m2a = 2b2 + 2c2 − a2 .
9.1.2 Important Points and Lines in a Triangle
• The circumcenter O is the center of the triangle’s circumcircle. It can be foundas the intersection of the perpendicular bisectors of the sides.
• The incenter I is the center of the incircle and thus the intersection of the anglebisectors.
• The orthocenter H is the intersection of the three altitudes of the triangle.• The centroid G is the intersection of the three medians.• The excenter JA is the center of the excircle lying in the angle BAC, outside the
triangleABC and tangent to the lines AB,AC and to the segment BC. There arethree excenters, JA, JB, JC , using similar definitions.
• If MA,MB,MC are the midpoints of BC,CA,AB respectively, then the threelines MAJA,MBJB,MCJC are concurrent, and their intersection point is calledthe mittenpunkt.
• If SA, SB, SC are the contact points of the excircles with BC,CA,AB respec-tively, then ASA,BSB,CSC are concurrent and their intersection is called theNagel point Na.
• Let TA, TB, TC be the contact points of the incircle with sides BC,CA,AB re-spectively. Then the lines ATA,BTB,CTC are concurrent, and their intersectionpoint is called the Gergonne point Ge.
• The isogonal conjugate of the point X is the intersection of the three concurrentlines that are the reflections ofAX,BX,CX about the angle bisectors ofA,B,Crespectively.
• The intersection point of the symmedians is called the symmedian point (or theLemoine point). It is the isogonal conjugate of the centroid G.
• Two lines passing through the vertex A of a triangle ABC are called isotomic ifthey cut the side BC in points symmetric with respect to the midpoint of BC.Given a point P , the isotomic lines of the Cevians AP , BP , and CP meet at apoint P ′ called the isotomic point of P .
• The Fermat point F of an acute triangle is the point that minimizes the sum of thedistances to the vertices. If we draw equilateral triangles BCA′, ABC′, ACB ′ onthe outside of the triangleABC, then the three linesAA′, BB ′, CC′ are concurrentat F .
• The Euler line: the points O,G,H are collinear and |HG| : |GO| = 2 : 1.
9.1 Compendium of Triangle Basic Terminology and Formulas 219
• The Gergonne point Ge, the triangle centroid G, and the mittenpunkt M arecollinear, with |Ge G| : |GM| = 2 : 1.
• The Nagel line: the points I,G,Na are collinear and |Na G| : |GI | = 2 : 1.
Kimberling tabulated and enumerated properties of more than 1477 centers in a trian-gle (later Brisse extended the list to 2001 items). The interested reader might consulthis updated web page:
• C. Kimberling: seehttp://faculty.evansville.edu/ck6/encyclopedia/ETC.html
9.1.3 Coordinates in a Triangle
We are given a reference triangle ABC.
Barycentric coordinates. The barycentric coordinates of the point P are given byan ordered triple FP = (tA, tB, tC) (up to nonzero scalar multiplication) such that Pis the centroid of the system consisting of the three masses tA, tB, tC located at therespective verticesA,B,C of the triangle, i.e., the mass tA is located atA, etc. Thesecoordinates were introduced by Möbius in 1827. The coordinates tA are proportionalto the areas area(PBC). If they are actually equal to these areas, they are calledhomogeneous barycentric coordinates.
Here are the coordinates of the principal points in a triangle.
• FG = (1, 1, 1)• FO = (
a2(b2 + c2 − a2
), b2
(a2 + c2 − b2
), c2
(a2 + b2 − c2
))
• FJA = (−a, b, c), FI = (a, b, c)
• FGe = ((p − b)(p − c), (p − c)(p − a), (p − a)(p − b))
• FNa = (p − a, p − b, p − c)
• FH = ((a2 + c2 − b2
)(a2 + b2 − c2
),(a2 + b2 − c2
)(b2 + c2 − a2
),(
a2 + c2 − b2)(b2 + c2 − a2
))
• FK = (a2, b2, c2
)
The line determined by the points (s1, s2, s2) and (t1, t2, t3) has the equation
det
⎛
⎝s1 s2 s3t1 t2 t3x1 x2 x3
⎞
⎠ = 0.
Trilinear coordinates. The trilinear coordinates of a point P are given by an orderedtriple of numbers proportional to the directed distance from P to the sides, up tomultiplication by a nonzero scalar. If we normalize them so that they give the distancesto the sides, then these are called the exact trilinear coordinates. Trilinear coordinatesare also known as homogeneous coordinates. They were introduced by Plücker in1835. For instance, the vertices have trilinear coordinates (1 : 0 : 0), (0 : 1 : 0), and(0 : 0 : 1).
220 9 Glossary
For most important points in a triangle, the trilinear coordinates are given bytriples of the form (f (a, b, c) : f (b, c, a) : f (c, a, b)), for some function f onthe sides a, b, c. For instance, we have fH (a, b, c) = cosB cosC, fG(a, b, c) = 1
a,
fO(a, b, c) = cosA, fI (a, b, c) = 1. For the Fermat point, we can take fF (a, b, c) =sin(A+ π
3 ).The homogeneous barycentric coordinates corresponding to (x : y : z) are
(ax, by, cz). The trilinear coordinates of the isogonal conjugate of (x : y : z) are( 1x, 1y, 1z
).
9.2 Appendix: Pell’s Equation
Pell’s equation is the Diophantine equation
x2 − dy2 = 1.
The first treatment of this equation was actually given by Lord William Brounckerin 1657, but his solution was erroneously ascribed to John Pell by Euler. Lagrangedeveloped the theory of continued fractions, which gives the actual method of findingthe minimal solution, and published the first proof in 1766. However, some methods(the cyclic method) were already known to the Indian mathematicians Brahmaguptaand Bhaskara in the twelfth century.General solutions. Pell’s equation has infinitely many solutions if d is not a perfectsquare and none otherwise. If (x0, y0) denotes its minimal solution (called also thefundamental solution) different from (1, 0), then all solutions (xn, yn) are obtainedfrom it by means of the recurrence relations
xn+1 = x0xn + dy0yn, yn+1 = y0xn + x0yn, x1 = x0, y1 = y0.
The fundamental solution. Thus the main point is finding the fundamental solution.This can be achieved using continued fractions. According to Lagrange, the continuedfraction of a quadratic irrational number (i.e., a real number satisfying a quadraticequation with rational coefficients that is not rational itself) is periodic after somepoint; in our case, even more can be obtained, namely
√d =
[a0, a1, . . . , am, 2a0
].
The numbersaj can be calculated recursively, as follows. Seta0 = [√d]
and constructthe sequences
P0 = 0, Q0 = 1,P1 = a0, Q1 = d − a2
0,
Pn = an−1Qn−1 − Pn−1, Qn = d−P 2n
Qn−1.
Then
an =[a0 + Pn
Qn
].
9.2 Appendix: Pell’s Equation 221
Furthermore, consider the sequences
p0 = a0, q0 = 1,p1 = a0a1 + 1, q1 = a1,
pn = anpn−1 + pn−2, qn = anqn−1 + qn−2.
Then we have the identities
p2n − dq2
n = (−1)n+1Qn+1,
which permit one to construct solutions to the Pell-type equations.If am+1 = 2a0, as happens for a quadratic irrational, thenQm+1 = 1. In particular,
we find that the fundamental solution is determined by the parity of m, as follows:
x0 = pm, y0 = qm, if m is odd,
x0 = p2m+1, y0 = q2m+1, if m is even.
Observe that pr/qr = [a0, a1, . . . , ar ]. The general solution (xn, yn) is obtained bymeans of the trick
x2 − dy2 = (x20 − dy2
0 )n = 1,
and setting
x + √dy = (
x0 + √dy0
)n, x − √
dy = (x0 − √
dy0)n,
we obtain the family of solutions
xn =(x0 + √
dy0)n + (
x0 − √dy0
)n
2, yn =
(x0 + √
dy0)n − (
x0 − √dy0
)n
2.
The negative Pell equation. The Pell-like equation
x2 − dy2 = −1
can also be solved, but it does not always have solutions. When it has one, then it hasinfinitely many. A necessary condition for this equation to be solvable is that all oddprime factors of d be of the form 4k + 1, and that d �≡ 0 (mod 4). However, theseconditions are not sufficient, as can be seen from the equation for d = 34, which hasno solutions. The method of continued fractions works for these equations as well. Infact, for this Pell-type equation, we have the fundamental solution
x0 = pm, y0 = qm, if m is even.
The equation does not have solutions if m is odd. Furthermore, the general solutionis given again by
xn =(x0 + √
dy0)n + (
x0 − √dy0
)n
2, yn =
(x0 + √
dy0)n − (
x0 − √dy0
)n
2,
222 9 Glossary
but only for odd n.
Pell-like equations. The Pell-like equation
x2 − dy2 = c
can be solved using the same ideas. If |c| < √d, then the equation has solutions if
and only if
c ∈{Q0,−Q1, . . . , (−1)m+1Qm+1
}.
Moreover, if c >√d , then the procedure is significantly more complicated. See for
instance the comments of Dickson on this subject. It is clear that using the solutions(xn, yn) to the companion Pell equation x2 −dy2 = 1 and a particular solution (u, v)of the general Pell-type equation from above, we are able to find infinitely manysolutions by means of the recurrence
un = xnu± dynv, vn = xnv ± ynu.
However, even if we start with the minimal solution, we cannot always obtain allsolutions of the equation using this recurrence. The reason is that we might well haveseveral fundamental solutions. For instance, if d = 10 and c = 9, then we have thefundamental solutions (7, 2), (13, 4) and (57, 18).
The general equationax2 − by2 = c
can be reduced to the former equation by setting d = ab, x′ = ax, c′ = ac andlooking for those solutions where x′ is divisible by a.
The degree-two equation in two unknowns. Moreover, the general degree-two equa-tion
ax2 + bxy + cy2 + dx + cy + f = 0
where a, b, c, d, e, f ∈ Z, can be reduced to either Pell-type equations or ellipticequations.
1. The case � = b2 − 4ac < 0. The equation represents an ellipse in the plane,and therefore it has only finitely many integer solutions. These equations were solvedby Gauss and earlier considered by Euler. According to Dickson one can find all itssolutions using the solutions to Pell’s equations.
2. If � = 0, the parabolic-type equation becomes
1
4a
((2ax + by + d)2 + 2(ae − bd)y + 4af − d2
)= 0.
If 2ae− bd = 0, then the equation is equivalent, through a linear transformation,to
(2ax + by + d)2 = d2 − 4af ;hence the parabola degenerates into two lines. If d2 − 4af is a perfect square, thenwe have two infinite families of solutions 2ax + by + d = ±√
d2 − 4af ; otherwise,there are no integer solutions.
9.2 Appendix: Pell’s Equation 223
If 2ae−bd �= 0, then we have to solve the equation 2(ae−bd)y = −t2+d2−4af ,for an arbitrary parameter t ∈ Z. This depends only on computing the residues modulo2(ae − bd) of the perfect squares, and yields infinitely many solutions if one exists.Furthermore, 2ax = t − d − by determines x.
3. If � > 0, then the hyperbolic-type equation becomes:
a(�x − 2cd + bc)2 + b(�x − 2cd + bc)(�y + 2ac + bd)+ c(�y − 2ac + bd)2
= −�(ac2 + cd2 + ef 2 − bdc − 4acf
),
which, through a transformation u = �x − 2cd + bc, v = �y − 2ac+ bd, becomes
au2 + buv + cv2 = h.
This equation can be reduced to a Pell equation. It will have then either zero orinfinitely many solutions.
The Ankeny–Artin–Chowla Conjecture. Although the Pell equation seems to bewell understood, there are many subtle questions concerning its solutions. Here is aproblem that has resisted to all efforts until now.
1. If p is a prime p ≡ 1 (mod 4) and y0 is the smallest positive value of y such that
x2 − py2 = −4,
then y0 �≡ 0 (mod p).2. If p is a prime p ≡ 3 (mod 4) and y0 is the smallest positive value of y such that
x2 − py2 = 1,
then y0 �≡ 0 (mod p).
The first conjecture is due to Ankeny, Artin, and Chowla, and the second to Mordell.
The Thue theorem for higher degree. The situation of equations of higher degreeis completely different. Let
f (x, y) = a0xn + a1x
n−1y + a2xn−2y2 + · · · + any
n
be a homogeneous form of degree n. If n ≥ 3 and f is irreducible over Q, then theequation
f (x, y) = g(x, y),
where g(x, y) is an arbitrary form of degree n − 2 (not necessarily homogeneous),has only finitely many solutions. This is a celebrated theorem due to Thue and basedon results of Siegel and Roth. Later A. Baker gave explicit bounds on the size of thesolutions for a wide class of such f .
• L.E. Dickson: Pell equation: ax2 + by2 + c made square, Chapter 12 in Historyof the Theory of Numbers, Vol. 2: Diophantine Analysis, New York, Chelsea,341–400, 1952.
• H.W. Lenstra, Jr.: Solving the Pell equation, Notices Amer. Math. Soc. 49 (2002),2, 182–192.
Index of Mathematical Results
Cebotarev density theorem, 83
Ankeny–Artin–Chowla conjecture, 223
Birch–Swinnerton-Dyer conjecture, 81Blaschke theorem, 211Bonnesen’s isoperimetric inequality, 170Bonsé–Pósa inequality, 46Borsuk problem, 153, 156, 175Borsuk–Lusternik–Shnirelman theorem, 175
Catalan’s conjecture, 56Cauchy’s integral formula, 167Cauchy’s rigidity theorem, 171, 173Cauchy–Schwartz inequality, 167Cauchy–Schwarz inequality, 175Ceva’s theorem, 139Chebyshev theorem, 40, 45, 46, 58Chinese Remainder theorem, 82Chinese remainder theorem, 50, 78congruent numbers problem, 80
De Moivre’s identity, 60Dirac theorem, 97Dirichlet theorem on arithmetic progressions,
50, 82, 97
Eisenstein criterion, 38Erdos conjecture, 114Erdos theorem, 126Erdos–Straus conjecture, 63Euler conjecture, 42Euler criterion, 72, 73Euler theorem in a triangle, 182
Fermat equation, 42, 67four color theorem, 104Frobenius coin problem, 50fundamental theorem of symmetric
polynomials, 85, 89
Gowers theorem, 123Green–Tao theorem, 123
Hadamard inequality, 186Hahn–Banach lemma, 131Heawood conjecture, 103Helly’s theorem, 159
intermediate value theorem, 188isoperimetric inequality, 147
Jensen inequality, 179Jung’s theorem, 159
Kahn–Kallai theorem, 156Kaprekar constant, 95
Lagrange four squares theorem, 74Lagrange mean value theorem, 194Lagrange multiplier method, 189Lagrange theorem on periodic continued
fractions, 220Laplace method, 198Lehmer conjecture, 57Ljunggren equation, 69
Maclaurin series, 189
226 Index of Mathematical Results
maximum principle for holomorphicfunctions, 211
Means inequality, 161, 179, 180, 189Millennium problems, 81, 195Minkowski’s theorem, 165
odd perfect number conjecture, 59
Pell equation, 43Pell’s equation, 220Picard theorem, 214pigeonhole principle, 86, 92, 96, 97, 108,
112
quadratic reciprocity theorem, 50
Rayleigh–Beatty theorem, 101Riemann hypothesis, 195
Roché inequality, 28, 182
Sturm comparison theorem, 212superperfect numbers conjecture, 60Sylvester’s problem, 172Szemerédi theorem, 122
Taylor formula, 209Thales’ theorem, 140Thue’s theorem, 223
Ulam problem, 126
Van der Waerden theorem, 122
Waring problem, 52, 74, 76Wiles theorem, 42Wilson theorem, 44, 53
Index of Mathematical Terms
abundant number, 9, 81altitude, 22, 25, 27, 139, 160, 162, 175, 217annulus, 190area, 8, 19, 23, 25–27, 70, 80, 124, 140–142,
145, 154, 157, 159, 161, 165–167, 170,173, 176, 177, 181, 182, 190, 217
arithmetic progression, 4, 7, 12, 40, 50, 56,82, 87, 97, 122
asymptotic, 16, 78, 106, 114, 128
ball, 25, 27, 143, 156–159, 174, 175Banach space, 214barycentric coordinates, 142, 219Bernoulli numbers, 54Bessel function, 210binomial
binomial coefficient, 14, 41, 62, 97, 98,105, 109
binomial expansion, 38, 62, 104binomial coefficient, 3, 37bisector, 21, 25, 27, 133, 134, 162, 175, 217body, 25, 145, 156–158, 166boundary, 18, 25, 150, 151, 156–158, 161,
165, 166, 174, 211
cardinal of a set, 15, 17, 50, 78, 88, 92,100–102, 104, 108, 111, 112, 117, 128,191
cardinality of a set, 15centroid, 218Cevian, 139choice number, 3, 37chord, 21, 22, 24, 135–137, 152, 157, 159,
160
circular permutation, 62, 190circumcenter, 218circumscribable, 23, 140collinear points, 23, 126, 138, 140–142, 172,
182, 218, 219commensurable, 126, 203compact, 34, 151, 157, 212–214complex number, 13, 24, 30, 89, 92, 126,
135, 136, 147, 148, 190, 200complex plane, 55computer search, 43, 63, 65computer-assisted proof, 104concave, 149congruence modulo p, 3, 9, 39, 40, 44, 50,
52, 61, 63, 66, 72, 73, 78–80, 82, 83congruence transformation, 24, 151congruent number, 80continued fraction, 220convergent, 9, 30–33, 77, 78, 190, 191, 193,
194, 203, 205convex, 16, 22–26, 106, 136, 139, 143, 144,
147, 149, 156–158, 167, 168, 170, 171convex hull, 136, 156, 158covering, 27, 173critical, extremal point, 180, 190
deficient number, 9, 81dense, 34, 126, 131, 142, 157, 203, 211density, 83, 109, 122, 128determinant, 11, 86, 87, 109, 138, 141, 186diameter, 24, 25, 145, 147, 151, 152, 154,
156, 158, 167, 174differential equation, 198digits, 4, 5, 13, 40, 47, 95
228 Index of Mathematical Terms
Diophantine equation, 42, 53, 56, 64, 67, 69,70, 80, 115, 220
discriminant, 48, 83, 194distance, 16, 19, 20, 22–24, 26, 107, 122,
126, 128, 130, 131, 136, 142, 143, 145,146, 148, 151, 152, 165, 167, 169, 219
domain, 26, 170, 171, 174
eigenvalue, 90, 91, 211Euler inequality, 28, 177Euler line, 22, 138, 218Euler totient function, 3, 6, 7, 39, 51, 57, 61,
92Euler–Poincaré characteristic, 103, 150, 171,
172Eulerian circuit, 19, 123excenter, 218
factorial, 3, 37Fermat point, 22, 138, 218Fibonacci sequence, 118, 185fractional part, 193function, 12, 23, 31, 33, 88, 144, 165, 194,
204, 207additive function, 208bijective function, 15, 102concave (or convex) function, 179continuous function, 33, 34, 71, 153, 157,
206, 209, 210decreasing function, 31, 139, 196entire function, 34, 213function, 104generating function, 16, 106holomorphic function, 211hyperbolic trigonometric function, 203increasing function, 15, 64, 101, 110, 189meromorphic function, 214periodic function, 31, 193positive function, 32, 198rational function, 13, 42, 92smooth function, 32–34, 87, 162, 198,
199, 208, 212
Gauss map, 156Gergonne point, 23, 141, 218, 219graph, 15, 19, 97, 102, 103, 123, 147, 150
Hadwiger number, 163Hamiltonian cycle, 97
homographic division, 138homothety, 26, 106, 126, 136, 170, 171
incenter, 218index of mathematical terms, 37infinite descent method, 56, 67inscribable, 8, 22, 23, 26, 70, 137, 140, 141,
169integer part, 4, 39integral equation, 34, 209integral operator, 211intrinsic perimeter, 164isosceles, 18, 21, 25, 122, 134, 135, 138,
159, 160isotomic point, 218
Jacobian matrix, 190
lattice, 26, 122, 129, 165Legendre symbol, 72, 82Lemoine (symmedian) point, 218length, 19, 21–27, 29, 94, 96, 124, 125, 135,
137, 140, 143, 145, 146, 151, 152,154–156, 159, 160, 162, 166–169,171, 175, 182, 187, 211, 217
linearlinear combination, 88, 93, 209linear equation, 113linear function, 33, 86, 204, 207linear functional, 131linear independence, 13, 25, 91, 92, 162linear map, 131
matrix, 26, 86, 87, 91, 109, 111, 141, 142,162, 164, 186, 190
median, 27, 175, 217Mersenne prime number, 55Mersenne primes, 59Minkowski norm, 164mittenpunkt, 218
Nagel line, 219Nagel point, 23, 141, 218nilpotent, 90, 93non-collinear points, 19noncollinear points, 27, 124, 126, 172nonlinear operator, 214number of combinations, 3, 37
orthocenter, 218
Index of Mathematical Terms 229
orthogonal, 130, 136, 137, 143, 145, 153–157
packing, 19, 125, 129parameterization (of a curve), 41partition, 15, 17, 26, 99, 103, 113, 114, 151,
153, 156, 158, 168, 169, 171, 174pentagonal number, 8, 69perfect number, 59periodic, 31, 193, 203, 225poder, 181polygon, 16, 23, 26, 29, 106, 121, 143, 144,
146, 147, 150, 158, 168, 171, 187polyhedron, 27, 153, 157, 171–173polynomial
characteristic polynomial, 91irreducible polynomial, 3, 38, 83, 92, 223monic polynomial, 34, 214polynomial with arbitrary coefficients, 11,
13, 86, 91, 93, 211, 214polynomial with integer coefficients, 6, 9,
12, 40, 51, 61, 82, 89real polynomial, 12, 29, 30, 33, 86, 87,
185, 188, 206, 208set of polynomials, 32, 197symmetric polynomial, 85, 89
projective duality, 172projective transformation, 138Pythagorean triple, 5, 42, 44, 66, 67, 69
quadratic irrational number, 220quadratic residue, 50, 68, 72, 73
�, 29, 185relatively prime, 39, 40, 46, 65, 78, 86, 93Riemann zeta function, 54, 194root, 34, 61, 83, 89, 91–93, 160, 200, 212
sequence, 4, 6, 9, 12, 14, 17, 18, 30–32, 40,43, 48, 50, 61, 74, 76, 77, 81, 87, 95,96, 98, 100, 110, 112, 114, 116, 117,119, 127, 128, 147, 191, 193, 196, 201,203, 206, 207, 213
series, 77simplex, 25, 156, 158, 159, 171stereographic projection, 41sum-free set, 17, 114superperfect number, 60surface, 26, 103, 154, 166, 174symmedian, 181symmetric, 137, 154, 155, 161, 166, 174, 175symmetry, 51, 88, 97, 106, 133, 136, 137,
160, 166, 170, 172, 205
tiling, 121, 149torus, 13, 93transcendental number, 55, 128triangular number, 8, 69trilinear coordinates, 138, 141, 219
vector space, 90, 130, 208
width, 23, 145, 152, 153, 155, 166, 167
Index of Topics and Methods
Classical arithmetic functions and ap-plications: 1.4, 1.29, 1.34, 1.36, 1.42,1.43, 1.44, 1.45, 1.46, 1.69
Congruences and divisibility: 1.1, 1.3,1.8, 1.11, 1.15, 1.17, 1.22, 1.24, 1.27,1.38, 1.48, 1.50, 1.51, 1.67, 1.68, 2.7, 2.45
Rational and algebraic numbers: 1.47,2.14, 2.15
Polynomials: 1.2, 1.7, 1.30, 2.2, 2.4, 2.9,2.12, 2.13, 2.24
Estimates of arithmetically definedfunctions: 1.5, 1.6, 1.23, 1.28, 1.37, 1.55,1.65, 1.66, 2.25, 2.35, 2.47
Symmetric functions: 2.1, 2.8
General Diophantine equations (Pyth-agorean, exponential): 1.12, 1.14, 1.16,1.20, 1.25, 1.33, 1.35, 1.39, 1.41, 1.49,1.52, 1.56, 1.57, 1.60
Pell’s equation and its applications:1.13, 1.58, 1.59
Continued fractions: 1.13
Prime numbers, Chebyshev theorem:1.9, 1.18, 1.19, 1.21, 1.32, 1.54, 2.46
Arithmetic progressions: 1.10, 1.26,1.40, 2.5, 2.21
Representing integers by algebraicfunctions, Waring problem: 1.31, 1.53,1.61, 1.62, 1.63, 1.64
Algebraic number theory: 1.70
Groups: 2.10
Combinatorial identities: 2.33, 2.34
Algebraic equations: 2.3, 2.16
Linear algebra, matrices, determi-nants: 2.6, 2.11, 2.17, 2.18, 2.40, 2.43
Combinatorial properties of sets of in-tegers: 2.26, 2.27, 2.28, 2.29, 2.41, 2.44
Sequences of integers: 2.19, 2.42, 2.52,2.53
Counting problems: 2.23, 3.21
Extremal problems concerning sets ofintegers: 2.36, 2.37, 2.48, 2.49, 2.50,2.51
232 Index of Topics and Methods
Ramsey-type problems: 2.20, 2.30,2.32, 2.38, 2.39, 2.56, 2.57, 2.59, 2.62
Graphs and applications: 2.22, 2.31,2.58
Tiling, packing and covering: 2.55,2.60, 2.64, 3.46, 3.47, 3.53
Area and volume: 3.42, 3.45, 3.51, 3.56,3.60
Distances in combinatorial geometry:2.61, 2.63, 3.23, 3.48, 3.54
Geometry of the triangle: 3.1, 3.11,3.12, 3.14, 3.16
Complex numbers and applications:3.5, 3.6, 3.8, 4.1, 4.19, 4.34
Geometric constructions, geometricloci: 3.2, 3.3, 3.4, 3.7, 3.9, 3.10, 3.15
Trigonometry: 3.17, 3.30
Analytic methods in geometry: 3.18
Higher dimensional geometry: 3.13,3.57
Convex geometry, integral formulas,isoperimetric inequalities: 3.24, 3.25,3.29, 3.49, 3.50, 3.55
Extremal problems concerning finitesets of points: 2.54, 3.20, 3.27, 3.28,3.31, 3.37, 3.40, 3.41, 3.44, 3.52
General geometric inequalities: 3.22,3.26, 3.43, 4.5
Geometric inequalities in the triangle:3.62, 3.63, 3.64, 3.65, 3.66, 3.67, 3.68,3.69, 3.70, 3.71, 3.72, 3.73
Topological methods in geometry: 3.32,3.33, 3.34, 3.35, 3.36, 3.38, 3.39, 3.58,3.59, 3.61
General inequalities: 4.2, 4.3, 4.4, 4.6,4.14, 4.15, 4.22, 4.26
Convexity of real functions: 4.16
Minima and maxima of real functions,mean values: 4.7, 4.8, 4.9, 4.10, 4.37
Functions of a complex variable, holo-morphic functions: 4.20, 4.24, 4.44,4.47
Sums of series and convergence of se-quences: 4.11, 4.12, 4.17, 4.18, 4.23,4.28, 4.29, 4.31
Dynamical systems, density of sets ofreals: 3.19, 4.32, 4.35, 4.43
Integrals: 4.13, 4.21, 4.30
Functional equations and inequations:4.27, 4.33, 4.36, 4.38
Integral and differential equations:4.25, 4.39, 4.40, 4.41, 4.42, 4.45
Linear functionals and operators: 2.65,4.46, 4.48
